Lakhmir Singh Class 10 Physics 1st Chapter “Electricity” solution

Lakhmir Singh Manjit Kaur Class 10 Physics 1st Chapter “Electricity” solution

Lakhmir Singh Physics solution: “Electricity” Chapter 1. Here you get easy solutions of Lakhmir Singh Physics solution Chapter 1 . Here we have given Chapter 1 all solution of Class 10. Its help you to complete your homework.

Board – CBSE
Text Book – Physics
Class – 10
Chapter – 01

Lakhmir Singh Class 10 Physics 1th Chapter Solution

(First Part)

Very short answer type questions:

1.) By what other name is the unit joule/ coulomb called?

Ans:

The other name of the unit joule/ coulomb is called as volt.

Because, 1 volt = 1J/ 1C.

2.) Which of the following statements correctly defines a volt?

a) a volt is a joule per ampere.

b) a volt is a joule per coulomb

Ans:-

The option b is correct.

Because a volt is a joule per coulomb.

3.) a) what do the letters p.d. stands for?

Ans:

The letters p.d. stands for potential difference.

b) which device is used to measure p.d.?

Ans:

To measure the potential difference between any two points voltmeter is used.

4.) What is meant by saying that the electric potential at a point is 1volt?

Ans:

The potential difference between any two points is said to be one volt when 1 joule of work is done to move the 1 coulomb of electric charge from one point to another point.

5.) How much work is done when one coulomb charge moves against a potential difference of 1 volt?

Ans:

1 joule of work is done when one coulomb charge moves against a potential difference of 1 volt.

6.) What is the SI unit of potential difference?

Ans:

The SI unit of potential difference is Volt and denoted by V.

7.) How much work is done in moving a charge of 2C across two points having a potential difference of 12V?

Ans:

We know that,

Potential difference = work done/ charge moved

V= W/Q

Hence, W = V*Q

Given that, V= 12V and Q= 2C

Thus, W= V*Q = 12*2= 24 J

Thus, 24J of work is done in moving a charge of 2C from one point to another with potential difference between them is 12V.

8.) What is the unit of electric charge?

Ans:

The unit of electric charge is coulomb and it is denoted by C.

9.) Define one coulomb charge?

Ans:

1 coulomb charge is the charge which is equivalent to the collective charge of 6.25*1018 number of electrons.

10.) Fill in the blanks with suitable words:

a) potential difference is measured in ______ by using a ______ placed in _______ across a component.

b) copper is a good _______ plastic is an _______.

Ans:

a) potential difference is measured in voltsby using a voltmeter placed in parallelacross a component.

b) copper is a good conductor; plastic is an insulator.

Short answer type questions:-

1.) What is meant by conductors and insulators? Give two examples of conductors and two of insulators.

Ans:

Conductors:

The materials or substances which allows flow of electric current through them are the conductors.
Conductors are having low resistivity.
Silver, copper and aluminium are best conductors of electricity.

Insulators:

Insulators are the materials or substances which does not allows the electric current through them.
Insulators are having high resistivity.
Glass, Ebonite, rubber, plastics etc are the insulators.

12.) Which of the following are conductors and which are insulators?

Sulphur, silver, copper, cotton, aluminium, air, nichrome, graphite, paper, porcelain, Mercury, mica, Bakelite, polythene, Manganin.

Ans:

Following are the conductors:

Silver, copper, aluminium, nichrome, Manganin, mercury.

Following are the insulators:

Cotton,air, paper, porcelain, mica, Bakelite, polythene, sulphur.

13.) What do you understand by the term electric potential? Or potential at a point? What is the unit of electric potential?

Ans:

Electric potential at a point is nothing but the amount of work done in bringing a unit positive charge from infinity to that point.
It is denoted by symbol V.
The SI unit of potential difference is volt.
That means 1V will be the potential difference at a point when 1J of work is done for moving a charge of 1C from infinity to that point.
Hence, it is given by

Potential difference = world one / quantity of charge moved

Hence, V= W/Q

14.) a) state relation between potential difference, work done and charge moved.

b) calculate the work done in moving a charge of 4C from a point at 220V to another point at 230 volt.

Ans:

The relation between potential difference, work done and charge moved is given by,
Potential difference= work done / quantity of charge moved
V= = W/ Q

Given that,

Q= 4C

V= 230-220= 10V

W= ?

Hence, work done in moving charge of 4C from a point at 220V to another point at 230V is given by,

We know that,

V= W/Q

Hence, W = V*Q = 10*4= 40J

15.) a) name a device that helps to measure the potential difference across a conductor.

b) how much energy is transferred by a 12V power supply to each coulomb of charge which moves around a circuit?

Ans:

a) voltmeter is used to measure the potential difference across a conductor.

b) we know that,

V= W/Q

This, W= V*Q

Given that,

V= 12V

Q= 1C

Thus, we can find work done in moving a 1 coulomb of charge as

W= V*Q = 12*1= 12J

Thus, the amount of work done in moving a charge of 1C is nothing but the energy transferred by a power supply to each coulomb of charge which it moves around the circuit.

Long answer type questions:

16.) a) what do you understand by the term potential difference?

Ans:

The potential difference is difference of electric potential between two points.
In other words, in an electric circuit the potential difference between two points is the amount of work done to move the unit positive charge from one point to the another.
And it is given by,

Potential difference= work done/ quantity of charge moved

V= W/Q

And the SI unit of potential difference is volt which is denoted by V.

b) what is meant by saying that the potential difference between two points is 1volt?

Ans:

We know that, potential difference is nothing but the amount of work done in moving charge from one point to other point.
Now we have to define one volt of potential difference between any two points.
Thus, potential difference between any two points in an electric circuit is said to be 1 volt only when 1 J of work is done in moving 1 C of charge from one point to another point.

c) what is the potential difference between the terminals of battery if 250J of work is required to transfer 20C of charge from one terminal of battery to other?

Ans:

Given that,

W = 250 J

Q= 20C

Then potential difference between two points is given by,

V= W/Q = 250/ 20 = 12.5V

d) what is a voltmeter? How is voltmeter is connected in circuit to measure the potential difference between two points. Explain with the help of diagram.

Ans:

The electrical instrument used to measure potential difference between any two points of an electric circuit is called as voltmeter.
Voltmeter is always connected in parallel across the two points whose potential difference we have to measure.
The following diagram shows the voltmeter is connected in parallel in an electric circuit to measure potential difference between two points.

e) state whether a voltmeter has a high resistance or low resistance. Give reason for your answer.

Ans: A voltmeter has a high resistance due to which it takes current from the circuit which is negligible.

(Second Part)

Electric Current

Very Short Answer Type Questions-

1) By what name is the physical quantity coulomb/second called?

Ans:- The physical quantity coulomb/ second is called as amperes.

Since, 1A = 1C/ 1S

2) What is the flow of charge called?

Ans:

The flow of charge is called as electric current.
Because, current = charge/ time

3) What actually travels through the wires when you switch on a light?

Ans: When we switch on the light then actually electrons will flow through the wires.

4) Which particles constitute the electric current in a metallic conductor?

Ans: Electrons are the charged particles which constitutes the electric current in a metallic conductor.

5) (a) In which direction does conventional current flow around a circuit?

(b) In which direction do electrons flow?

Ans:

Current flows from positive terminal of battery to the negative terminal of battery through the outer circuit.
Thus, the direction of flow of electric current is opposite to the direction of flow of electrons.

b) Electrons flows from negative terminal of battery to the positive terminal of battery i.e. opposite to the direction of flow of conventional current.

6) Which of the following equation shows the correct relationship between electrical units?

1 A = 1 C/s or 1 C = 1 A/s

Ans:

1A= 1C/1S, is the correct relationship between current, charge and time i.e. electrical units.

7) What is the unit of electric current?

Ans:

The SI unit of electric current is ampere and it is denoted by A.
Since, current = charge/time
Hence, 1A= 1C/1S

8) (a) How many milliamperes are there in 1 ampere?

(b) How many microamperes are there in 1 ampere?

Ans:

Since 1mA = 10^-3A
Hence, 1A = 10³ mA = 1000A

Since, 1uA= 10^-6A
Hence, 1A= 10⁶ uA

9) Which of the two is connected in series: ammeter or voltmeter?

Ans: Ammeter is connected in series with the circuit to measure the electric current flowing through it.

10) Compare how an ammeter and a voltmeter are connected in a circuit.

Ans:

Ammeter is an electrical instrument used to measure electric current through the circuit.
And ammeter is always connected in series with the circuit.
Voltmeter is an electrical instrument which is connected in parallel with the circuit to measure the potential difference between any two points of the circuit.

11) What do the following symbols mean in circuit diagrams?

Ans:

a) Symbol a show the variable resistance or rheostat. By using this changing value of resistance, we can change the current without changing applied voltage.

b) The symbol b shows that the closed switch means circuit is closed and current is flowing through circuit.

12) If 20 C of charge pass a point in a circuit in 1 s, what current is flowing?

Ans:

Given that,

Q= 20C

Time t= 1s

We know that, current is the rate of flow of charge i.e. electrons flowing through the circuit.

Current= charge/ time

I= 20/1 = 20A

13) A current of 4 A flows around a circuit for 10 s. How much charge flows past a point in the circuit in this time?

Ans:

Given that,

Current= 4A

Time t= 10s

Charge=?

We know that, the rate of flow of charge is the current flowing through the circuit.

Hence, current= charge/time

I= Q/t

Q= I*t = 4*10= 40C

Thus, 40C of charge flows through the circuit for 10s when current is 4A.

14) What is the current in a circuit if the charge passing each point is 20 C in 40 s?

Ans:

Given that,

Charge = 20C

Time = 40s

We know that, the rate of flow of charge is called as the current flowing through the circuit.

Hence, current = charge / time

Current= 20/40= 1/2= 0.5A

15) Fill in the following blanks with suitable words:

(a) A current is a flow of……………..For this to happen there must be a ………..circuit.

(b) Current is measured in…………..using an…………..placed in…………in a circuit.

Ans:

a) A current is a flow of electrons. For this happen there must be a closed circuit.

b) Current is measured in ampere using an ammeter placed in series in a circuit.

Short Answer Type Questions

16) (a) Name a device which helps to maintain potential difference across a conductor (say, a bulb).

(b) If a potential difference of 10 V causes a current of 2 A to flow for 1 minute, how much energy is transferred?

Ans:

a) The device used to maintain potential difference across a conductor say bulb is the cell or battery.

b) Potential difference = 10V

Current = 2A

Time = 1min = 60s

We know that, the energy transferred is given by,

E= VIt = 10*2*60= 1200J

17) (a) What is an electric current? What makes an electric current flow in a wire?

(b) Define the unit of electric current (or Define ampere).

Ans:

Electric current is the rate of flow of electrons through the wire or conductor when a cell or battery is applied across its two ends.
When we connect a cell or battery of specific voltage across the ends of the conducting wire then there will be an electric force is acting on the electrons which are present in the wire.
Due to the electric force acting on electrons they move from negative terminal of battery to the positive terminal of battery, since electrons are negatively charged particles.
Thus, these flow of electrons through the wire constitutes current which in the opposite direction of flow of electrons i.e., from positive terminal of battery to the negative terminal of battery as shown if figure below.

The SI unit of electric current is ampere.
When there is flow of 1C electric charge through the conducting wire for 1s then the electric current flowing through that conducting wire should be 1 A.
Thus, 1A= 1C/ 1s

18) What is an ammeter? How is it connected in a circuit? Draw a diagram to illustrate your answer.

Ans:

Ammeter is an electric instrument which is used to measure the electric current flowing through the electric circuit.
And ammeter is always connected in series with the circuit in which we have to measure the electric current.
The following figure shows the ammeter is connected in series with a conductor AB to measure electric current flowing through it.

19) (a) Write down the formula which relates electric charge, time and electric current.

(b) A radio set draws a current of 0.36 A for 15 minutes. Calculate the amount of electric charge that flows through the circuit.

Ans:

Following is the formula which gives the relationship between electric charge Q, time t and electric current I.
I= Q/t
Since, electric current is the rate of flow of electric charge flowing through the circuit.

Given that,

I= 0.36 A

Time t= 15 min = 15*60= 900 seconds

Thus, the amount of electric charge flowing through the circuit is given by,

I= Q/ t

Hence, Q= I*t = 0.36*900= 324C

Thus, the amount of electric charge flowing through the circuit will be 324 coulombs.

20) Why should the resistance of

(a) an ammeter be very small?

(b) a voltmeter be very large?

Ans:

The resistance of an ammeter is low because the entire current is passing through the ammeter and hence it doesn’t change the value of an electric current flowing through the circuit.

We know that, voltmeter is used to measure the potential difference between two points in the electrical circuit.
The resistance of voltmeter is large so that it takes the negligible current from the circuit.

21) Draw circuit symbols for

(a) fixed resistance

(b) variable resistance

(d) a battery of three cells

(e) an open switch

(f) a closed switch.

Ans:

Following figure shows the symbols:

22) What is a circuit diagram? Draw the labelled diagram of an electric circuit comprising of a cell, a resistor, an ammeter, a voltmeter and a closed switch (or closed plug key). Which of the two has a large resistance: an ammeter or a voltmeter?

Ans:

Circuit diagram shows how the different electrical components are connected in an electrical circuit in the form of their symbols and electric circuit is the continuous closed diagram.
The following figure shows the labelled diagram of an electric circuit consisting of a cell, a resistor, an ammeter, a voltmeter and a closed switch.
Voltmeter has large resistance due to which it takes negligible current from the circuit.

23) If the charge on an electron is 1.6 × 10–19 coulombs, how many electrons should pass through a conductor in 1 second to constitute 1 ampere current?

Ans:

We know that, when 1 coulomb of electric charge flows through the electrical circuit for 1 second then 1 ampere of current will flow through the circuit.
Given that, 1.6*10^-19 C is the charge on each electron,
Hence, we have to find the number of electrons in 1coulomb.
Hence,
Number of electrons in 1C = 1/1.6*10^-19 = 0.625*10^-19 = 6.25*10¹⁸
Thus, to flow 1ampere of current through the electrical circuit, the number of electrons flowing through it will be 6.25*10¹⁸.

24) The p.d. across a lamp is 12 V. How many joules of electrical energy are changed into heat and light when?

(a) a charge of 1 C passes through it?

(b) a charge of 5 C passes through it?

(c) a current of 2 A flows through it for 10 s?

Ans:

Given that,

Potential difference = 12V

a) Charge = 1C

We know that, the electrical energy is given by,

E= P*t= V*I*t

But, I= Q/t

So, Q= I*t

Thus, E= 12*1= 12J

b) Charge Q= 5C

We know that,

E= VIt = V*Q = 12*5 = 60 J

c) Current I= 2A

Time t= 10 s

Hence, electrical energy is given by,

E= VIt = 12*2*10= 240J

25) In 10 s, a charge of 25 C leaves a battery, and 200 J of energy are delivered to an outside circuit as a result.

(a) What is the p.d. across the battery?

(b) What current flows from the battery?

Ans:

Given that,

Time t= 10s

Charge Q= 25 C

Energy E = 200J

a) We know that,

E = VIt

Hence, V= E/It = 200/Q = 200/25 = 40/5= 8 V

Hence, the potential difference across the battery will be 8V.

b) As we know that, electric current is the rate of flow of electric charge.

Hence, I= Q/t = 25/ 10 = 2.5 A

Thus, 2.5 A current is flowing from the battery.

Long Answer Type Question

26) (a) Define electric current. What is the SI unit of electric current?

(b) One coulomb of charge flows through any cross-section of a conductor in 1 second. What is the current flowing through the conductor?

(c) Which instrument is used to measure electric current? How should it be connected in a circuit?

(d) What is the conventional direction of the flow of electric current? How does it differ from the direction of flow of electrons?

(e) A flash of lightning carries 10 C of charge which flows for 0.01 s. What is the current? If the voltage is 10 MV, what is the energy?

Ans:

The electric current is the rate of flow of electric charge or electrons flowing through the electrical circuit.
The SI unit of electric current is ampere and it is denoted by A.

As given, that, 1C of charge flows through any cross section of conductor in 1 second then the current flowing through the electrical circuit will be 1 ampere.
Since, 1A = 1C/ 1s

An electrical instrument ammeter is used to measure an electric current flowing through the electrical circuit.
Ammeter is always connected in series with the circuit in which current is to be measured.

The conventional direction of flow of electric current is from positive terminal of battery to the negative terminal of battery.
And the direction of flow of electrons is from negative terminal of the battery to the positive terminal of the battery.
Hence, the conventional direction of current is opposite to the direction of flow of electrons.

Given that,

Charge Q= 10C

Time t= 0.01 s

Voltage V= 10MV = 10*106 = 107 V

Hence, current flowing through it is given by,

I= Q/t = 10/0.01 = 1000 A

Electrical energy is given by,

E = VIt =107* 103* 0.01= 108 J

E= 100*106 J

E= 100 MJ

(Third Part)

Ohm’s Law

Very Short Answer Type Questions –

1) Name the law which relates the current in a conductor to the potential difference across its ends.

Ans: The law which relates the current in a conductor to the potential difference across its ends is the Ohm’s law.

2) Name the unit of electrical resistance and give its symbol.

Ans: The SI unit of electrical resistance is ohm and it is denoted by the symbol Ω.

3) Name the physical quantity whose unit is “ohm”.

Ans: The physical quantity having SI unit as ohm is the electrical resistance.

4) What is the general name of the substances having infinitely high electrical resistance?

Ans: The substances having infinitely high resistance are called as insulators.

5) Keeping the resistance constant, the potential difference applied across the ends of a component is halved. By how much does the current change?

Ans:

Given that, resistance is kept constant and the potential difference applied across the ends of the component is halved then how much current will be flows.
We know that,
According to Ohm’s law,

V= IR

But here R is constant.

Hence,

If potential difference V is halved then electric current, I becomes also get halved.

6) State the factors on which the strength of electric current flowing in a given conductor depends.

Ans:

According to Ohm’s law,

V= IR

Thus, I = V/R

Hence, the strength of the electrical current depends on the potential difference V and the resistance R.

7) Which has less electrical resistance: a thin wire or a thick wire (of the same length and same material)?

Ans:

Given that, there are two wires of the same length and same material but one is thin wire and the other is thick wire.
As we know that, resistance of material is directly proportional to its length and inversely proportional to its area of cross section.
If the wire is thinner its area of cross section decreases. so electrical resistance get increased.
And as the wire is thick its area of cross section increases and hence resistance get decreased.
Hence, the thick wire has less resistance.

8) Keeping the potential difference constant, the resistance of a circuit is halved. By how much does the current change?

Ans:

Given that, the potential difference is kept constant and resistance of the circuit is halved.
We know that, according to Ohm’s law,

V= IR

Thus, I= V/R
Here, V is kept constant and R is halved then current become doubled.

9) A potential difference of 20 volts is applied across the ends of a resistance of 5 ohms. What current will flow in the resistance?

Ans:

Given that,

Voltage V= 20V

Resistance R= 5 ohm

I=?

We know that,
According to Ohm’s law,

V= IR

So, I= V/R = 20/5 = 4A
Hence, the current flowing through the resistance will be 4A.

10) A resistance of 20 ohms has a current of 2 amperes flowing in it. What potential difference is there between its ends?

Ans:

Given that,

Resistance R= 20 ohm

Current I= 2A

We know that,

According to Ohm’s Law,

V= IR

V= 20*2= 40V

Thus, the potential difference between its ends will be 40V.

11) A current of 5 amperes flows through a wire whose ends are at a potential difference of 3 volts. Calculate the resistance of the wire.

Ans:

Given that,

Current I= 5A

Potential difference V= 3V

R=?

We know that,
According to Ohm’s law,

V= IR

R= V/I = 3/5 = 0.6 ohm

Thus, the resistance of the wire will be 0.6 ohm

12) Fill in the following blank with a suitable word

Ohm’s law states a relation between potential difference and …………………..

Ans: Ohm’s law states a relation between potential difference and current.

Short Answer Type Questions

13) Distinguish between good conductors, resistors and insulators. Name two good conductors, two resistors and two insulators.

Ans:

On the basis of the electrical resistance of the materials they are classified as good conductors, resistors and insulators.

Good conductors:

Good conductors are those materials which are having very low electrical resistance.
As the resistance of good conductors is very low, they allow electricity to flow through them easily.
Silver, copper and aluminium are the good conductors.

Resistors:

The substances having high electrical resistance are called as resistors.
Resistors reduces the electric current in the circuit.
The alloys of nichrome, Manganin and constantan are the resistors.

Insulators:

The materials having high resistance are called as insulators.
As the resistance of insulators is high, they do not allow electricity through them.
Rubber and wood are the insulators.

14) Classify the following into good conductors, resistors and insulators:

Rubber, Mercury, Nichrome, Polythene, Aluminium, Wood, Manganin, Bakelite, Iron, Paper, Thermocol, Metal coin

Ans:

Following materials are the good conductors:

Aluminium, iron, metal coin, mercury

Following materials are the resistors:

Nichrome, Manganin,

Following materials are the insulators:

Rubber, wood, polythene, Bakelite, thermocol, paper.

15) What is Ohm’s law? Explain how it is used to define the unit of resistance.

Ans:

Ohm’s Law:

According to Ohm’s Law, at constant temperature the current flowing through the electrical conductor is directly proportional to the potential difference between its two ends.

I V

Hence, V I

V= RI

Where R is constant which is called as resistance of the conductor.

Hence, R = V/I

Where V is the voltage measured in volt.

And I is the current measured in ampere.

Hence, R = volt/ ampere

Thus, 1ohm = 1V/ 1A

Hence, the SI unit of resistance is ohm.

16) (a) What is meant by the “resistance of a conductor”? Write the relation between resistance, potential difference and current.

(b) When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Calculate the value of the resistance of the resistor.

Ans:

a) Resistance is the property of an electrical conductor due to which it opposes the flow of current through it.

The relation between resistance R, potential difference V and electric current I is given by,

V= IR

b) Given that,

Voltage V= 12V

Current I= 2.5mA = 2.5*10-3 A

R=?

We know that, according to Ohm’s law,

V= IR

So, R = V/I = 12/2.5*10-3 = 4.8*10³ ohm = 4800 ohm

17) (a) Define the unit of resistance (or define the unit “ohm”).

(b) What happens to the resistance as the conductor is made thinner?

(c) Keeping the potential difference constant, the resistance of a circuit is doubled. By how much does the current change?

Ans: – a)

We know that, according to Ohm’s Law,

V= IR

R= V/I

Hence, 1ohm = 1V/ 1A

Thus, 1 ohm is the resistance of conductor when there is a potential difference of 1 volt is applied to it and current flowing through it is 1 ampere.

b) We know that, the resistance of conductor is directly proportional to its length and inversely proportional to its area of cross section. As the conductor is made thinner its area of cross section decreases and hence resistance of the conductor increases.

c) Given that, the potential difference is kept constant and the resistance of circuit is doubles.

We know that,

V= IR

I = V/ R

Here V is constant and R is doubled, hence the current flowing through the circuit will be halved.

18) (a) Why do electricians wear rubber hand gloves while working with electricity?

(b) What p.d. is needed to send a current of 6 A through an electrical appliance having a resistance of 40 W?

Ans:

a) Electricians wear rubber hand gloves while working with electricity because rubber is having very high resistance and hence rubber is the insulator and protects from the electric shocks.

b) Given that,

Current I = 6A

Resistance R = 40 ohm

We know that, according to Ohm’s law,

V= IR

V= 6*40 = 240 V

Hence, the potential difference needed will be 240V.

19) An electric circuit consisting of a 0.5 m long nichrome wire XY, an ammeter, a voltmeter, four cells of 1.5 V each and a plug key was set up.

(i) Draw a diagram of this electric circuit to study the relation between the potential difference maintained between the points ‘X’ and ‘Y’ and the electric current flowing through XY.

(ii) Following graph was plotted between V and I value.

What would be the values of V I ratios when the potential difference is 0.8 V, 1.2 V and 1.6 V respectively?

What conclusion do you draw from these values?

(iii) What is the resistance of the wire?

Ans:

a) The following diagram shows the circuit diagram to study the relation between potential difference maintained between the two points X and Y and electric current flowing through XY.

b) As the V-I graph is a straight line passing through origin hence the slope of VI graph is constant.

Here, slope of VI graph is, V/I = resistance.
Thus, the resistance of the circuit is constant.

c) From graph,

At I = 0.2 A, V = 0.5V

Hence, R = 0.5/0.2 = 5/2 = 2.5 ohm

Thus, the resistance of the wire will be 2.5 ohm

Long Answer Type Question

20) (a) What is the ratio of potential difference and current known as?

(b) The values of potential difference V applied across a resistor and the corresponding values of current I flowing in the resistor are given below. Plot a graph between V and I, and calculate the resistance of the resistor.

Potential difference (V)	2.5	5.0	10.0	15.0	20.0	25.0
Current (A)	0.1	0.2	0.4	0.6	0.8	1.0

(c) Name the law which is illustrated by the above V–I graph.

(d) Write down the formula which states the relation between potential difference, current and resistance.

(e) The potential difference between the terminals of an electric iron is 240 V and the current is 5.0 A. What is the resistance of the electric iron?

Ans:

a) The ratio of potential difference and the electric current is known as the resistance of the circuit.

The following diagram shows the VI graph for given values of V and I.
And the resistance found will be 25 ohms.

c) The above VI graph illustrates the Ohm’s law.

d) The formula which gives the relationship between potential difference, current and resistor is given by,

V= IR

e) Given that,

V = 240 V

I= 5A

R=?

We know that, according to Ohm’s law,

V= IR

R = V/ I = 240/5 = 48 ohm

Thus, the electrical resistance of the electric iron will be 48 ohm.

Factors affecting the resistance of the conductor:

Very Short Answer Type Questions

1) What happens to the resistance as the conductor is made thicker?

Ans: If the conductor is made thicker then resistance of the conductor decreases.

2) If the length of a wire is doubled by taking more of wire, what happens to its resistance?

Ans: If the length of the wire is doubled by taking more wire, then it’s resistance also get doubled because resistance of the wire is directly proportional to the length of the wire.

3) On what factors does the resistance of a conductor depend?

Ans: The resistance of the conductor depends on the following factors:

Length of the conductor
Area of cross section of the conductor

4) Name the material which is the best conductor of electricity.

Ans: Silver metal is the best conductor of electricity as it has lowest resistivity.

5) Which among iron and mercury is a better conductor of electricity?

Ans: Iron is the better conductor of electricity among mercury and iron. Because resistivity of iron is less than the resistivity of mercury.

6) Why are copper and aluminium wires usually used for electricity transmission?

Ans: Copper and aluminium wires usually used for electricity transmission because they are having low resistivity and cheaper to use as compared to silver metal which is the best conductor of electricity.

7) Name the material which is used for making the heating element of an electric iron.

Ans: Nichrome is the alloy which is used for making heating element of an electric iron.

8) What is nichrome? State its one use.

Ans:

Nichrome is a alloy having high very high resistivity and it does not undergo oxidation easily even at high temperature.
Nichrome alloy is used as heating element in electric iron, toaster, electric Kittles, room heaters, water heaters and in hair dryers.

9) Give two reasons why nichrome alloy is used for making the heating elements of electrical appliances.

Ans:

Nichrome alloy is used for making the heating elements of electrical appliances because,
Nichrome is a alloy having high very high resistivity and it does not undergo oxidation easily even at high temperature.

10) Why are the coils of electric irons and electric toasters made of an alloy rather than a pure metal?

Ans: The heating elements of electrical appliances such as electric iron and toaster etc are made up of alloy rather than pure metal because the alloy has resistivity which is higher than the pure metal and also alloy does not undergo oxidation easily even at high temperature when it is red hot.

11) Which has more resistance –

(a) a long piece of nichrome wire or a short one?

(b) a thick piece of nichrome wire or a thin piece?

Ans:

a) A long piece of nichrome wire has more resistance.

b) Thin piece of nichrome wire has more resistance.

12) (a) How does the resistance of a pure metal change if its temperature decreases?

(b) How does the presence of impurities in a metal affect its resistance?

Ans: a) If the temperature of pure metal is decreased then it’s resistance also decreases.

b) The presence of impurities in a metal increases the resistance of the metal.

13) Fill in the following blanks with suitable words

Resistance is measured in………… The resistance of a wire increases as the length………; as the temperature………….; and as the cross-sectional area………….

Ans: Resistance is measured in ohms. The resistance of a wire increases as the length increases; as the temperature increases; and as the cross-sectional area decreases.

Short Answer Type Questions :-

14) (a) What do you understand by the “resistivity” of a substance?

(b) A wire is 1.0 m long, 0.2 mm in diameter and has a resistance of 10 W. Calculate the resistivity of its material?

Ans:

As we know that, the resistance of conductor is directly proportional to length of the conductor and inversely proportional to the area of cross section of the conductor then we can write as

R α l/A

Hence, R= ql/A

Where q is the constant of proportionality called as resistivity of the conductor.
It is also called as specific resistance of the conductor.
And the resistivity of conductor is numerically equal to the resistance of conductor whose length is 1meter and area of cross section is 1meter square.

b) Given that,

Length of the wire = 1m

Diameter = 0.2mm

Radius of the wire = 0.1mm = 10^-4 m

Resistance of wire = 10 ohm

The resistivity of the material is given by,

q= RA/l = (10*3.14*10^-8)/1

= 3.14*10^-7 ohm meter

15) (a) Write down an expression for the resistance of a metallic wire in terms of the resistivity. (b) What will be the resistance of a metal wire of length 2 metres and area of cross-section 1.55 × 10–6 m2, if the resistivity of the metal be 2.8 × 10–8 W m?

Ans: a) If l is the length of the metallic wire, A is the cross section and q is the resistivity of the wire then resistance of the wire R is given by,

R= q*l/A

b) Length of wire = 2m

Area of cross section A= 1.55*10^-6 m2

Resistivity of the metal = 2.8*10^-8 ohm meter

The resistance of the metal wire is given by,

R= q*l/A

= 2.8*10^-8*2/ 1.55*10^-6

= 5.6*10^-2/1.55= 3.61*10^-2 ohm

= 0.0361 ohm

16) (a) Give two examples of substances which are good conductors of electricity. Why do you think they are good conductors of electricity?

(b) Calculate the resistance of a copper wire 1.0 km long and 0.50 mm diameter if the resistivity of copper is 1.7 × 10–8 Wm.

Ans:

a) Silver, copper, aluminium are the good conductors of electricity as they are having low resistivity.

b) Given that,

Length of copper wire= 1km = 10³m

Diameter= 0.5mm

Radius of wire = 0.25mm = 0.25*10^-3 m

Resistivity of wire = 1.7*10^-8 ohm meter

Then the resistance of the wire is given by,

R= q*l/A

= (1.7*10^-8*10³)/(3.14*0.0625*10^-6)

= (1.7*10^-5)/(0.19625*10^-6)

= 17/0.19625

= 86.62 ohm

17) Will current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Give reason for your answer.

Ans:

The current will flow more easily through a thick wire of the same material when connected to the same source than thin wire.
Because the thick wire has less resistance than thin wire.

18) How does the resistance of a conductor depend on

(a) length of the conductor?

(b) area of cross-section of the conductor?

(c) temperature of the conductor?

Ans:

a) The resistance of conductor is directly proportional to the length of the conductor.

b) The resistance of conductor is inversely proportional to the area of cross section of the conductor.

c) Resistance of the conductor is directly proportional to the temperature of the conductor.

19) (a) Give one example to show how the resistance depends on the nature of material of the conductor.

(b) Calculate the resistance of an aluminium cable of length 10 km and diameter 2.0 mm if the resistivity of aluminium is 2.7 × 10–8 Wm.

Ans:

If we have taken two wires having same length and same diameter but one wire is made up of copper metal and other is made up of nichrome alloy.
And we observed that, the resistance of nichrome wire is more as compared to the copper wire even if they are having same length and same area of cross section.
This proves that, the resistance of the conductor depends on the nature of the material of the conductor.

b) Length of aluminium cable = 10km = 10*10³ m

Diameter = 2mm

Radius of cable = 1mm = 10^-3 m

Resistivity= 2.7*10^-8 ohm meter

Then resistance of the wire is given by,

R= q*l/A

=(2.7*10^-8*10⁴)/(3.14*10^-6)

= 2.7*10²/ 3.14

= 0.8598*10² ohm

= 85.98 ohm

= 86 ohm

20) What would be the effect on the resistance of a metal wire of

(a) increasing its length?

(b) increasing its diameter?

(c) increasing its temperature?

Ans: a) As we know that, the resistance of the conductor is directly proportional to its length hence if the length of the conductor is increased its resistance also increases.

b) As we know that, the resistance of the conductor is inversely proportional to the area of cross section of the conductor hence if the diameter of the conductor is increased then it’s radius also increases due to which area of cross section increases and hence resistance of the conductor decreases.

c) As we know that, resistance of conductors of pure metal is directly proportional to the temperature of the conductor hence if the temperature of the metal of conductor is increased then it’s resistance also increases.

21) How does the resistance of a wire vary with its

(a) area of cross-section?

(b) diameter?

Ans:

a) As we know that, the resistance of conductor is inversely proportional to the area of cross section of the conductor hence if the area of cross section increases the resistance of the conductor decreases.

As we know that, resistance of conductor is inversely proportional to the area of cross section of the conductor which means resistance of conductor is also inversely proportional to the diameter of the conductor.
And hence if the diameter of the is increased its resistance becomes decreases.

22) How does the resistance of a wire change when

(i) its length is tripled?

(ii) its diameter is tripled?

(iii) its material is changed to one whose resistivity is three times?

Ans:

As we know that, resistance of conductor is directly proportional to its length.
If the length of the conductor is tripled then it’s resistance also get tripled.

b) If the diameter of the conductor is tripled then it’s resistance becomes 1/9th times its original resistance.

c) If the material is changed to one whose resistivity is three times then also the resistance of the conductor become three times. Because resistance of the conductor is directly proportional to the resistivity of the conductor.

23) Calculate the area of cross-section of a wire if its length is 1.0 m, its resistance is 23 W and the resistivity of the material of the wire is 1.84 × 10–6 W m.

Ans: Given that,

Length = 1m

Resistance = 23 ohm

Resistivity = 1.84*10^-6 ohm meter

We know that,

R= qL/A

A = q*l/R = 1.8*10^-6*1/23

= 0.07826 *10^-6 m2

= 7.8*10^-8 m2

Long Answer Type Question

24) (a) Define resistivity. Write an expression for the resistivity of a substance. Give the meaning of each symbol which occurs in it.

(b) State the SI unit of resistivity.

(c) Distinguish between resistance and resistivity.

(d) Name two factors on which the resistivity of a substance depends and two factors on which it does not depend.

(e) The resistance of a metal wire of length 1 m is 26 W at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature?

Ans:

The resistivity of the material of conductor is numerically equal to the resistance of the conductor of that material whose length is one meter and area of cross section is 1m2.
The expression for resistivity of the conductor is given by,

q= R*A/l

Where, q is the resistivity of the material of conductor

R is the resistance of the conductor

A is the area of cross section of the conductor

L is the length of the conductor.

b) The SI unit of resistivity is ohm meter.

Resistance of the material of conductor is the opposition to the flow of electrons.
While resistivity of the conductor is the resistance of conductor having specific dimensions.
Resistance is denoted by R and resistivity is denoted by q.
The SI unit of resistance is ohm while the SI unit of resistivity is ohm meter.

Resistance is given by,

R= q*l/A

And resistivity is given by,

q= R*A/l

Resistance depends on the temperature, length and cross-sectional area of conductor.
While resistivity depends on the temperature and nature of the material of conductor.

The resistivity of the substance depends on the nature of the material of the conductor and temperature of the conductor.
The resistivity of the substance does not depend on the length of the conductor and area of cross section of the conductor.

Given that,

Length = 1m

Resistance = 26ohm

Diameter = 0.3mm

Radius = 0.15mm= 0.15*10^-3 m

The resistivity of the wire is given by,

q= R*A/l

= (26*3.14*0.15*0.15*10^-6)/ 1

= 1.83*10^-6 ohm meter

Combination of Resistors

Very Short Answer Type Questions

1) Give the law of combination of resistances in series.

Ans: The law of combination of resistances in series states that the resultant resistance of number of resistors connected in series is equal to the sum of their individual resistance.

2) If five resistances, each of value 0.2 ohm, are connected in series, what will be the resultant resistance?

Ans: Given that 5 resistances each having resistance 0.2 ohm are connected in series then their resultant resistance is given by,

Re= 0.2 + 0.2 + 0.2 + 0.2 + 0.2 = 5*0.2 = 1ohm

Thus, their equivalent resistance will be 1 ohm.

3) State the law of combination of resistances in parallel.

Ans: According to law of resistances is parallel, when number of resistances are connected in parallel then their equivalent resistance is less than the smallest individual resistance of the resistor connected in parallel.

4) If 3 resistances of 3 ohm each are connected in parallel, what will be their total resistance?

Ans:

Given that, 3 resistances each having resistance 3 ohm are connected in parallel then their equivalent resistance is given by,

1/Re = 1/3 + 1/3 + 1/3 = 3/3 = 1

Re = 1 ohm

Thus, their equivalent resistance will be 1 ohm.

5) How should the two resistances of 2 ohms each be connected so as to produce an equivalent resistance of 1 ohm?

Ans:

Here, the equivalent resistance given is 1 ohm which is less than the individual resistance 2 ohm of 2 resistances.
Hence, we connect these 2 resistors in parallel to get equivalent resistance 1 ohm.

Thus,

1/Re = 1/2 + 1/2 = 1

Re= 1 ohm

6) Two resistances X and Y are connected turn by turn:

(i) in parallel, and (ii) in series. In which case the resultant resistance will be less than either of the individual resistances?

Ans:

Given that, the two resistances X and Y are connected in parallel and in series.
Then their equivalent resistance will be less than either of the individual resistance only in parallel combination.

7) What possible values of resultant resistance one can get by combining two resistances, one of value 2 ohm and the other 6 ohm?

Ans:

Given that, the two resistors of resistances 2 ohm and 6 ohm.
If we connect them in series then equivalent resistance will be,

Re = 2+ 6 = 8 ohm

If these same resistances are connected in parallel then their equivalent resistance will be,

1/Re = 1/2 + 1/6 = 4/6 = 2/3

Re = 3/2 = 1.5 ohm

8) Show how you would connect two 4 ohm resistors to produce a combined resistance of (a) 2 ohms (b) 8 ohms.

Ans:

a) To get 2 ohm as equivalent resistance we connect two resistors having resistance 4 ohm in parallel.

1/Re = 1/4 + 1/4 = 2/4 = 1/2

Re= 2 ohm

b) To get equivalent resistance as 8 ohm we connect two resistors having resistance 4 ohm in series.

Thus, Re = 4 + 4 = 8 ohm

9) Which of the following resistor arrangement, A or B, has the lower combined resistance?

Ans:

In figure A) the resistance is 10 ohm

And in figure B) resistances 10 ohm and 1000 ohm are connected in parallel combination then their equivalent resistance will be,

1/Re = 1/10 + 1/1000

= 101/1000

Re= 1000/101 = 9.9 ohm

Thus, in figure B) the combined resistance will be lower.

10) A wire that has

resistance R is cut into two equal pieces. The two parts are joined in parallel. What is the resistance of the combination?

Ans:

Given that, a wire of resistance R is cut into two equal parts.
Then resistance of each part will be R/2.
Now these two parts are joined in parallel combination then their equivalent resistance will be,

1/Re = 2/R + 2/R = 4/R

Re= R/4

Thus, their combined resistance in parallel combination will be R/4 ohm.

11) Calculate the combined resistance in each case:

Ans:

a) In figure a) resistances of 500 ohm and 1kohm are connected in series hence their combined resistance will be,

Re= 500 + 1000 = 1500 ohm

b) In figure b) two resistances of 2 ohm are connected in parallel then their combined resistance will be,

1/Re = 1/2 + 1/2

= 1

Re= 1 ohm

c) In figure c) two resistors of resistances 4 ohm are connected in parallel and to this parallel combination one resistor of resistance 3 ohm is connected in series.

Thus, their combined resistance will be,

Re = [(4*4)/4+4] + 3

= (16/8) + 3

= 2 + 3

= 5 ohm

12) Find the current in each resistor in the circuit shown below:

Ans:-

From figure,

The voltage across 6 ohm resistor is 24 ohm and hence current through 6ohm resistance is given by,

V= IR

I = V/R = 24/6= 4A

And the voltage across the 4 ohm resistor is also 24 volt and hence current flowing through 4 ohm resistance is given by,

V= IR

I = V/R = 24/4 = 6 A

Short Answer Type Questions –

13) Explain with diagrams what is meant by the “series combination” and “parallel combination” of resistances. In which case the resultant resistance is:

(i) less, and

(ii) more, than either of the individual resistances?

Ans:

Series combination:

When we connect two or more resistances end to end successively then these combination is said to be series combination of resistors as shown in figure.

According to law of series combination of resistors, the equivalent resistance of the any number of resistances connected in series is equal to the sum of their individual resistance.
If three resistors R1, R2 and R3 are connected in series combination then their equivalent resistance will be,

Re= R1 + R2 + R3

Parallel combination:

When two resistors are connected in between the two same points then this combination is called as parallel combination of resistors as shown in figure.

According to law of parallel combination, the reciprocal of the equivalent resistance of number of resistors connected in parallel is equal to the sum of reciprocals of all the individual resistances.
If the resistors R1, R2 and R3 are connected in parallel then the reciprocal of their equivalent resistance is given by,

1/Re = 1/R1 + 1/R2 + 1/R3

In parallel combination of resistors, the resistance will be less.
And in series combination of resistors the equivalent resistance will be more.

14) A battery of 9 V is connected in series with resistors of 0.2 W, 0.3 W, 0.4 W, 0.5 W and 12 W. How much current would flow through the 12 W resistor?

Ans: Given that,

Voltage =9V

Resistors connected in series: 0.2ohm, 0.3ohm, 0.4ohm, 0.5ohm, 12ohm.
The equivalent resistance when resistors are connected in series is given by,

Re = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 ohm

All the resistors are connected in series with voltage of battery is 9V.
And we know that, in series combination of resistors the current through each resistor will be same and it is given by,

I= V/ Re = 9/ 13.4 = 0.67A

Hence, current flowing through resistor 12 ohm is also 0.67A.

15) An electric bulb of resistance 20 W and a resistance wire of 4 W are connected in series with a 6 V battery. Draw the circuit diagram and calculate:

(a) total resistance of the circuit.

(b) current through the circuit.

(d) potential difference across the resistance wire.

Ans:- The following diagram shows an electric bulb of resistance 20 ohm and resistance wire of resistance 4ohm are connected in series with a 6V battery.

a) The total resistance of the circuit is given by,

R= 20 + 4 = 24 ohm

b) The current through the circuit is given by,

I = V/ R = 6/24 = 1/4 = 0.25 A

c) The potential difference across the electric bulb is given by,

V= IR = 0.25* 20 = 5 V

d) The potential difference across the resistance wire is given by,

V= IR = 0.25*4 = 1V

16) Three resistors are connected as shown in the diagram. Through the resistor 5 ohm, a current of 1 ampere is flowing.

(i) What is the current through the other two resistors?

(ii) What is the p.d. across AB and across AC?

(iii) What is the total resistance?

Ans:-

a) The voltage between two points B and C is given by,

V= IRe = 1*Re

Here,

1/Re = 1/10 + 1/15

1/Re = 25/150

= Re= 150/25 = 6 ohm

Hence, V= 1*6= 6V

Now, current flowing through resistor 10 ohm is given by,

I= V/R = 6/10= 3/5= 0.6 A

And current flowing through resistor 15 ohm is given by,

I= V/R = 6/15 = 2/5 = 0.4 A

b) The potential difference across AB is given by,

V= IR = 1*5= 5V

And potential difference across AC is given by,

V= I*(5+Re)

= 1*(5+6)

= 11V

c) The total resistance is given by,

R= 5 + Re

= 5+6

= 11 ohm

17) For the circuit shown in the diagram below:

What is the value of:

(i) current through 6 W resistor?

(ii) potential difference across 12 W resistor?

Ans:

18) Two resistors, with resistances 5 Ω and 10 Ω respectively are to be connected to a battery of emf 6 V so as to obtain:

(i) minimum current flowing

(ii) maximum current flowing

(a) How will you connect the resistances in each case?

(b) Calculate the strength of the total current in the circuit in the two cases.

Ans:

To get the minimum current flowing through the circuit we connect the resistances in series combination, since in series combination the equivalent resistance is more.
Hence, in series combination,

Re= 5+10= 15 ohm

Hence, minimum current flowing through the circuit is given by,

I= V/Re

= 6/15= 2/5= 0.4A

When we connect the resistances in parallel combination then only maximum current is flowing through the circuit, since in parallel combination the equivalent resistance is less.

Hence,

1/Re = 1/5 + 1/10 = 15/50

Re= 50/15 = 3.33 ohm

Hence, maximum current flowing through the circuit is given by,

I= V/Re

= 6/3.33

= 1.8 A

19) The circuit diagram given below shows the combination of three resistors R1, R2 and R3:

Find :(i) total resistance of the circuit.

(ii) total current flowing in the circuit.

(iii) the potential difference across R1.

Ans:

The total resistance of the circuit is given by,

R= 4 + [(3*6)/(3+6)]

R= 4 + 18/9 = 4 + 2 = 6 ohm

The total resistance of the circuit is 6 ohm.

b) The total current flowing through the circuit is given by,

I= V/R = 12/6 = 2A

c) The potential difference across R1= 4ohm is given by,

V= IR₁ = 2*4 = 8V

20) In the circuit diagram given below, the current flowing across 5 ohm resistor is 1 amp. Find the current flowing through the other two resistors.

Ans:

The voltage between two points AB is given by,

V= IR = 1*5= 5V

Thus, the current flowing through 4 ohm resistor is given by,

I= V/R = 5/4 = 1.25 A

And current flowing through the 10ohm resistor is given by,

I= V/R = 5/10= 0.5 A

21) A resistor has a resistance of 176 ohms. How many of these resistors should be connected in parallel so that their combination draws a current of 5 amperes from a 220 volt supply line?

Ans:

Given that,

Voltage supply = 220V

Current = 5A

The total resistance of the circuit becomes,

V= IR

R= V/I= 220/5= 44 ohm

Now, the equivalent resistance of resistors each having resistance 176 ohm in parallel combination is 44ohm.

Thus,

Let, n number if resistors having resistance 176 ohm are connected in parallel to get equivalent resistance as 44ohm.

Then,

1/44 = 1/176 + 1/176 + … n times

1/44= n/176

Hence, n = 176/44 = 4

Thus, 4 resistors having resistance 176 ohm will be connected in parallel combination.

22) An electric heater which is connected to a 220 V supply line has two resistance coils A and B of 24 W resistance each. These coils can be used separately (one at a time), in series or in parallel. Calculate the current drawn when:

(a) only one coil A is used.

(b) coils A and B are used in series.

(c) coils A and B are used in parallel.

Ans:

When only one coil is used then current flowing through circuit is given by,

I= V/R = 220/24 = 9.16 A

When coils A and B having resistance 24 ohm each are connected in series then current through the circuit is given by,

I= V/ R = 220/(24+24) = 220/48= 4.583 A

When coils A and B are used in parallel then current flowing through the circuit is given by,

I= V/Re = 220/Re

Here,

1/Re = 1/24 + 1/24 = 2/24 = 1/12

Re= 12 ohm

Thus,

I = 220/ 12 = 18.33 A

23) In the circuit diagram given below five resistances of 10 W, 40 W, 30 W, 20 W and 60 W are connected as shown to a 12 V battery.

Calculate: (a) total resistance in the circuit.

(b) total current flowing in the circuit.

Ans:

Here resistances 10 ohm and 40 ohm are connected in parallel then their equivalent resistance will be,

1/ Re = 1/10 + 1/40 = 50/400

Re = 400/50 = 8 ohm

And resistors 30 ohm, 20ohm and 60ohm are connected in parallel then their equivalent resistance is given by,

1/Re = 1/30 + 1/20 + 1/60

1/Re = 6/60

Re = 60/6 = 10 ohm

Hence, the total resistance of the circuit is given by,

R= 8 + 10 = 18ohm

The current flowing through the circuit is given by,

V= IR

I = V/R = 12/18= 0.67 A

24) In the circuit diagram given below, three resistors R1, R2, and R3 of 5 W, 10 W and 30 W, respectively are connected as shown.

Calculate: (a) current through each resistor.

(b) total current in the circuit.

(c) total resistance in the circuit.

Ans:

The current flowing through resistor 5ohm is given by,

I= V/R = 12/5 = 2.4 A

The current flowing through the resistor 10 ohm is given by,

I= V/R = 12/10= 1.2 A

The current flowing through the resistor 30 ohm is given by,

I= V/R = 12/30= 0.4A

The total current flowing through the circuit is given by,

I= 2.4 + 1.2 + 0.4 = 4A

Since, total resistance is given by,

1/Re = 1/5 + 1/10 + 1/30

1/Re = (6+3+1)/30

1/Re = 10/30

Re = 30/10 = 3 ohm

Thus, the total current flowing through the circuit is given by,

I= V/R = 12/3 = 4A

total resistance is given by,

1/Re = 1/5 + 1/10 + 1/30

1/Re = (6+3+1)/30

1/Re = 10/30

Re = 30/10 = 3 ohm

25) A p.d. of 4 V is applied to two resistors of 6 W and 2 W connected in series. Calculate: (a) the combined resistance

(b) the current flowing

(c) the p.d. across the 6 W resistor

Ans:

Given that,

Voltage = 4V

Resistors connected is series: 6ohm and 2ohm

Combined resistance is given by,

R= 6+2= 8 ohm

The current flowing through the circuit is given by,

I= V/R = 4/8= 0.5A

The potential difference across the resistor of 6ohm is given by,

V= IR = 0.5*6= 3V

26) A p.d. of 6 V is applied to two resistors of 3 W and 6 W connected in parallel. Calculate: (a) the combined resistance

(b) the current flowing in the main circuit

(c) the current flowing in the 3 W resistor.

Ans:

Given that,

Voltage= 6V

Resistors connected in parallel: 3ohm and 6ohm

The combined resistance is given by,

1/Re = 1/3 + 1/6 = (2+1)/6 = 3/6 = 1/2

Re= 2 ohm

The current flowing through the circuit is given by,

I= V/R = 6/2= 3A

The current flowing through the resistor of 3ohm is given by,

I= V/R = 6/3 = 2A

27) In the circuit shown below, the voltmeter reads 10 V.

(a) What is the combined resistance?

(b) What current flows?

(c) What is the p.d. across 2 Ω resistor?

(d) What is the p.d. across 3 Ω resistor?

Ans:

Given that,

Voltage = 10V

Resistor connected in series: 2ohm and 3ohm

The combined resistance will be,

Re= 2+3= 5ohm

Current flowing through the circuit is given by,

I= V/R = 10/5= 2A

The potential difference across 2ohm resistor is given by,

V= IR = 2*2= 4V

The potential difference across 3ohm resistor is given by,

V= IR = 2*3= 6V

28) In the circuit given below:

(a) What is the combined resistance?

(b) What is the p.d. across the combined resistance?

(c) What is the p.d. across the 3 Ω resistor?

(d) What is the current in the 3 Ω resistor?

(e) What is the current in the 6 Ω resistor?

Ans:

The combined resistance of the circuit is given by,

1/Re = 1/3 + 1/6 = (2+1)/6 = 3/6 = 1/2

Re= 2 ohm

The potential difference across the combined resistance will be given by,

V= IRe = 6*2= 12V

As the resistors 3 ohm and 6 ohm are connected in parallel the potential difference across both the resistors will be same.

Hence, potential difference across the resistor 3 ohm is 12V.

The current flowing through the 3ohm resistor is given by,

I= V/R = 12/3 = 4A

The current flowing through the resistor 6ohm is given by,

I= V/R = 12/6= 2A

29) A 5 V battery is connected to two 20 W resistors which are joined together in series.

(a) Draw a circuit diagram to represent this. Add an arrow to indicate the direction of conventional current flow in the circuit.

(b) What is the effective resistance of the two resistors?

(c) Calculate the current that flows from the battery.

(d) What is the p.d. across each resistor?

Ans:

The following circuit diagram shows the 5V battery is connected to two 20 ohm resistors which are joined together in series.

The effective resistance of the two resistors connected in series is given by,

Re= 20 + 20 = 40 ohm

The current flowing through the circuit is given by,

I= V/R = 5/40= 1/8 = 0.125A

The potential difference across each resistor of resistance 20 ohm is given by,

V= IR = 0.125*20= 2.5V

30) The figure given below shows an electric circuit in which current flows from a 6 V battery through two resistors.

(a) Are the resistors connected in series with each other or in parallel?

(b) For each resistor, state the p.d. across it.

(c) The current flowing from the battery is shared between the two resistors. Which resistor will have bigger share of the current?

(d) Calculate the effective resistance of the two resistors.

(e) Calculate the current that flows from the battery.

Ans:

The resistors are connected in parallel.

As the resistors are connected in parallel hence the potential difference across each resistor is the same and it is equal to the voltage applied.

Hence, voltage across each resistance is 6V.

The current flowing through the circuit is divided due to resistors in parallel.

And the resistor having lower resistance draws the more current.

Hence, resistor of 2ohm have bigger share of current.

The effective resistance of the two resistors is given by,

1/Re = 1/2 + 1/3 = 5/6

Re= 6/5 = 1.2 ohm

The current flowing through the circuit is given by,

I= V/R = 6/ 1.2 = 5 A

31) A 4 Ω coil and a 2 Ω coil is connected in parallel. What is their combined resistance? A total current of 3 A passes through the coils. What current passes through the 2 Ω coil?

Ans:

Given that,

The combined resistance is given by,

1/Re = 1/4 + 1/2 = 3/4

Re= 4/3 = 1.33 ohm

Given that, total 3A current is flowing through the circuit.

Hence, V= IRe = 3*1.33= 4V

The current flowing through 2ohm coil is given by,

I= V/R = 4/2= 2A

Long Answer Type Questions

32) (a) With the help of a circuit diagram, deduce the equivalent resistance of two resistances connected in series.

(b) Two resistances are connected in series as shown in the diagram:

(i) What is the current through the 5 ohm resistance?

(ii) What is the current through R?

(iii) What is the value of R?

(iv) What is the value of V?

Ans:

Let us consider the two resistors of resistances R1, R2 are connected in series as shown in figure.

Let I be total current flowing through the circuit and V be the voltage given.

We know that, in series combination the current flowing is same through all the resistors and voltage get divided at each resistor.

Let V1, V2 be the voltages across resistors R1, R2 respectively.

Hence, we can write

V= V1 + V2

According to Ohm’s law,

V= IR

Similarly,

V1= IR1, V2= IR2

Thus,

V= V1 + V2

IR= IR1 + IR2

IR= I*(R1 + R2)

R = R1 + R2

This is the derivation for equivalent resistance when two resistors are connected in series combination.

In series combination, the equivalent resistance is greater than each individual resistance connected in series.

The current flowing through the circuit is given by,

I= V/R = 10/5= 2A

Hence, resistance of second resistor is given by,

V= IR

R= V/I = 6/2= 3ohm

The equivalent resistance of two resistors connected in series is given by,

Re= 5+3= 8 ohm

The current flowing through 5ohm resistor is given by,

I= V/R = 10/5= 2A

Current through resistor R is also 2A, since in series combination the current flowing through each resistor is same.

The value of R here is 3 ohm.

The value of V here is

V= 10+6 = 16 V

33) (a) With the help of a diagram, derive the formula for the resultant resistance of three resistors connected in series.

(b) For the circuit shown in the diagram given below:

Calculate:

the value of current through each resistor.
the total current in the circuit.
the total effective resistance of the circuit.

Ans:

The following is the derivation for equivalent resistance when three resistors of resistances R1, R2 and R3 are connected in parallel combination.

Let us consider the three resistors of resistances R1, R2 and R3 are connected in series as shown in figure.

Let I be total current flowing through the circuit and V be the voltage given.

We know that, in series combination the current flowing is same through all the resistors and voltage get divided at each resistor.

Let V1, V2 and V3 be the voltages across resistors R1, R2 and R3.

Hence, we can write

V= V1 + V2 + V3

According to Ohm’s law,

V= IR

Similarly,

V1= IR1, V2= IR2, V3=IR3

Thus,

V= V1 + V2 + V3

IR= IR1 + IR2 + IR3

IR= I*(R1 + R2 + R3)

R = R1 + R2 + R3

This is the derivation for equivalent resistance when three resistors are connected in series combination.

In series combination, the equivalent resistance is greater than each individual resistance connected in series.

The current flowing through resistor 5ohm is given by,

I= V/R = 6/5 = 1.2 A

The current flowing through the resistor 10 ohm is given by,

I= V/R = 6/10 = 0.6A

The current flowing through the resistor 30 ohm is given by,

I= V/R = 6/30 = 1/5= 0.2A

The total current in the circuit is given by,

I= 1.2 + 0.6 + 0.2 = 2A

Also,

In given circuit diagram, resistors 5ohm, 10 ohm and 30 ohm are connected in parallel.

Then their equivalent resistance will be given by,

1/Re = 1/5 + 1/10 + 1/30

1/Re = (6+3+1)/30 = 10/30

Re = 30/10 = 3 ohm

Hence, current flowing through the circuit is given by,

I= V/ Re = 6/3= 2A

In given circuit diagram, resistors 5ohm, 10 ohm and 30 ohm are connected in parallel.

Then their equivalent resistance will be given by,

1/Re = 1/5 + 1/10 + 1/30

1/Re = (6+3+1)/30 = 10/30

Re = 30/10 = 3 ohm

34) (a) With the help of a circuit diagram, obtain the relation for the equivalent resistance of two resistances connected in parallel.

(b) In the circuit diagram shown below, find:

(i) Total resistance.

(ii) Current shown by the ammeter A

Ans:

Following is the derivation for equivalent resistance when two resistors R1 and R2 are connected in parallel combination.

Let us consider the two resistances R1 and R2 are connected in parallel as shown in figure.

V be the voltage applied between point A and B and I be the current flowing through the circuit.

Let I1 be the current flowing through the resistor R1 and I2 be the current flowing through the resistor R2.

Hence, total current flowing through the circuit is given by,

I = I1 + I2

According to Ohm’s law,

V= IR

Hence, I = V/R

And

I1 = V/R1 , I2= V/R2

Therefore,

I= I1 + I2

V/R = V/R1 + V/R2

1/R = 1/R1 + 1/R2

This is the equivalent resistance when two resistors of resistances R1 and R2 are connected in parallel.

In parallel combination the voltage across each resistor is same and current flowing through each resistor is different.

And the equivalent resistance is less than the smallest individual resistance.

The total resistance of the circuit is given by,

1/Re = 1/5 + 1/(3+2) = 1/5 + 1/5 = 2/5

Re = 5/2 = 2.5 ohm

The current flowing through the circuit is given by,

I= V/Re = 4/ (5/2) = 8/5 = 1.6 A

35) (a) Explain with the help of a labelled circuit diagram, how you will find the resistance of a combination of three resistors of resistances R1, R2 and R3 joined in parallel.

(b) In the diagram shown below, the cell and the ammeter both have negligible resistance. The resistors are identical.

With the switch K open, the ammeter reads 0.6 A. What will be the ammeter reading when the switch is closed?

Ans:

The following is the explanation of parallel combination of three resistors R1, R2 and R3 and their equivalent resistance.

Let us consider three resistors of resistances R1, R2 and R3 are connected in parallel combination as shown in figure.

Let V be the voltage across each resistor.

And I be the total current flowing through the circuit.

We know that, in parallel combination of resistors the voltage across each resistor is same but current get divided in parallel combination.

Let I1, I2 and I3 be the current flowing through the resistors R1, R2 and R3 respectively.

Then, total current is given by,

I= I1 + I2 + I3

According to Ohm’s law,

V= IR

I= V/R

Similarly,

I1= V/R1, I2= V/R2, I3= V/R3

Thus, I= I1 + I2 + I3

V/R = V/R1 + V/R2 + V/R3

V/R = V*(1/R1 + 1/R2 + 1/R3)

1/R = 1/R1 + 1/R2 + 1/R3

This is the derivation for equivalent resistance when three resistors are connected in parallel combination.

In parallel combination, the equivalent resistance is less than the smallest individual resistance.

When the switch K is open then only two resistors are in the circuit.

At that time ammeter shows reading 0.6A.

Hence, total resistance of the circuit when switch K is open is given by,

Let R be the resistance of each resistor.

Then,

1/R’ = 1/R + 1/R = 2/R

Thus, R’ = R/2

Now, we can find the potential difference of the cell,

V= IR’= 0.6*R/2 = 0.3R

But, when switch K is closed then there will be three resistors are in the circuit, at that time equivalent resistance will be,

1/R” = 1/R + 1/R + 1/R = 3/R

R” = R/3

Now, by Ohm’s law,

V= IR”

0.3R= I*R/3

0.3*3= I

I= 0.9 A

Thus, when the switch K is closed the ammeter will shows reading as 0.9 A.

Domestic Electric Circuits: Series or Parallel

Very Short Answer Type Questions

1) Are the lights in your house wired in series?

Ans:

Lights in our house are not wired in series.

2) What happens to the other bulbs in a series circuit if one bulb blows off?

Ans:

In a circuit with electric bulbs in series combination, if any one of the bulb blows off then all the other bulbs also stops glowing.

3) What happens to the other bulbs in a parallel circuit if one bulb blows off?

Ans:

In parallel combination of electric bulbs in electric circuit, if any one of the bulb glows off then also other bulbs are glowing.

4) Which type of circuit, series or parallel, is preferred while connecting a large number of bulbs:

(a) for decorating a hotel building from outside?

(b) for lighting inside the rooms of the hotel?

Ans:

For decorating a hotel building from outside series circuit is preferred.

For lighting inside the rooms of hotel parallel combination circuit is preferred.

5) Draw a circuit diagram to show how two 4 V electric lamps can be lit brightly from two 2 V cells.

Ans:

The following diagram shows the given circuit diagram.

Short Answer Type Questions

6) Why is a series arrangement not used for connecting domestic electrical appliances in a circuit?

Ans:

Because, in series combination the current drawn by each appliance will be same but the voltage drop across each electrical appliance will be different.
And when any one of the electrical appliances connected in series get stopped then all the other appliances will be stopped. Due to which series arrangement is not used for connecting domestic electrical appliances in a circuit.

7) Give three reasons why different electrical appliances in a domestic circuit are connected in parallel.

Ans:

The total effective resistance of resistors connected in parallel combination becomes low or less than the resistances connected.
The current through each component in parallel combination is also different.
In parallel combination, if any one of the components get damaged then the whole circuit not get damaged. In parallel combination we can make one component ON and other OFF at a time.
Now, if the parallel combination is used in domestic wiring, then voltage dropped across each electrical device will be same which is equal to the voltage applied.
Due to parallel combination, the effective resistance will be low and hence maximum current will be obtained. So that each device can draw a current requiring to it. Hence, the current flowing through each branch will be different due to parallel combination.
Also, if we want to turn OFF any one device and others have to make ON then we have to use parallel combination only. Also, if any one of the devices get damaged due to extra current then that device will be stopped only and whole circuit will be working, this is possible only in parallel combination.
Because of the all above advantages, the parallel combination is used in domestic wiring.

8) Ten bulbs are connected in a series circuit to a power supply line. Ten identical bulbs are connected in a parallel circuit to an identical power supply line.

(a) Which circuit would have the highest voltage across each bulb?

(b) In which circuit would the bulbs be brighter?

(c) In which circuit, if one bulb blows out, all others will stop glowing?

(d) Which circuit would have less current in it?

Ans:

Given that, ten bulbs are connected in series circuit to a power supply line. And ten identical bulbs are connected in parallel to the same power supply line.

The parallel combination of bulbs has the highest voltage across each bulb.

In parallel combination, the bulbs are glowing brighter.

In series combination of bulbs, if one bulb glows off then all other bulbs also glow off.

In series combination of bulbs circuit there will be less current is flowing through the circuit.

9) Consider the circuits given below:

(a) In which circuit are the lamps dimmest?

(b) In which circuit or circuits are the lamps of equal brightness to the lamps in circuit (i)?

(c) Which circuit gives out the maximum light?

Ans:

In second circuit diagram, the bulbs are connected in series combination and hence they are glowing dimmest.

In circuit first only one bulb is in the circuit.
And the bulbs in third circuit are connected in parallel combination which glows with the same brightness as the bulbs in first circuit.

As all the bulbs in third circuit glows brightest and with same light intensity hence bulbs in third circuit gives out maximum light.

10) If you were going to connect two light bulbs to one battery, would you use a series or a parallel arrangement ? Why ? Which arrangement takes more current from the battery?

Ans:

To connect two bulbs with one battery we use parallel combination.
Because, in parallel combination the voltage across each bulb is same and due to which it glows brightly.
And as the resistance in parallel combination is low hence these arrangement takes more current from the battery.

Long Answer Type Question

11) (a) Which is the better way to connect lights and other electrical appliances in domestic wiring: series circuits or parallel circuits ? Why?

(b) Christmas tree lamps are usually wired in series. What happens if one lamp breaks?

(c) An electrician has wired a house in such a way that if a lamp gets fused in one room of the house, all the lamps in other rooms of the house stop working. What is the defect in the wiring?

(d) Draw a circuit diagram showing two electric lamps connected in parallel together with a cell and a switch that works both lamps. Mark an A on your diagram to show where an ammeter should be placed to measure the current.

Ans:

The better way to connect lights and other electric appliances in domestic wiring in parallel combination.
Because, in parallel combination the voltage across each electrical appliance is same and if any one of the electric appliances get stopped working then also other appliances also works properly.

In Christmas tree all lights are connected in series combination and if any one of the lights get breaks, then all the lights also stop glowing.

Given that, an electrician has connected number of lamps in a house such that if any one of the lamps stops glowing in one room, then the lamps in other room also stops glowing.
Which means he connected lamps in series combination.

Following is the circuit diagram showing two electrical lamps connected in parallel together with a cell.

Electric Power

Very Short Answer Type Questions:

1) State two factors on which the electrical energy consumed by an electrical appliance depends.

Ans:

The electric energy consumed by an electrical appliance depends on two factors:
Power rating of the appliance
Time for which the appliance is used.

2) Which one has a higher electrical resistance: a 100 watt bulb or a 60 watt bulb?

Ans:

We know that, an electric power is inversely proportional to the resistance.
Hence, smaller is the resistance larger is the power and vice versa.
Given two bulbs are of electric power 100 watt and 60 watt.
Hence, 60 watt bulb has more resistance.

3) Name the commercial unit of electric energy.

Ans:

The commercial unit of electrical energy is kilowatt-hour.

4) An electric bulb is rated at 220 V, 100 W. What is its resistance?

Ans:

Given that,

V= 220 V

Power = 100W

The electric power is given by,

P= V²/R

Hence, resistance R is given by,

R= V²/P = 220*220/100= 484 ohm

5) What is the SI unit of (i) electric energy, and (ii) electric power?

Ans:

The SI unit of electrical energy is joule.
And the SI unit of electric power is joule per second.

6) Name the quantity whose unit is (i) kilowatt, and (ii) kilowatt-hour.

Ans:

Kilowatt is the commercial unit of electrical power.
And kilowatt hour is the commercial unit of electrical energy.

7) Which quantity has the unit of watt?

Ans:

The electrical power has unit watt.

8) What is the meaning of the symbol kWh? Which quantity does it represent?

Ans:

The symbol KWh represents the kilowatt-hour.
One Kilowatt-hour is the amount of energy consumed when an electrical appliance having power rating 1kilowatt is used for 1 hour.
And 1KWh= 36,00,000 joules.

9) If the potential difference between the end of a wire of fixed resistance is doubled, by how much does the electric power increase?

Ans:

If the potential difference between the two ends of the conductor is doubled then the power of the electric power becomes get increased four times.
Since, P= V²/R

10) An electric lamp is labelled 12 V, 36 W. This indicates that it should be used with a 12 V supply. What other information does the label provide?

Ans:

Given that,

V= 12V

P= 36 W

Indicates that an electric lamp is used with a 12V supply.

An electric lamp has power of 36W means it consumes energy at the rate of 36J per second.

11) What current will be taken by a 920 W appliance if the supply voltage is 230 V?

Ans:

Given that,

P= 920W

V= 230V

We know that, an electric power is given by,

P= VI

I= P/V = 920/230= 4A

Short Answer Type Questions

12) Define watt. Write down an equation linking watts, volts and amperes.

Ans:

The SI unit of electric power is watt and denoted by W.
The power of an electric appliance is said to be one watt only when an electric appliance consumes electrical energy at the rate of 1 joule per second.
An electric power in terms of voltage V and current I is given by,
P= VI
Hence, 1watt = 1volt*1 ampere

13) Define watt-hour. How many joules are equal to 1 watt-hour?

Ans:

The unit of an electrical energy is watt-hour.
When an electrical appliance of 1watt power is used for 1 hour then the amount of electrical energy consumed will be one watt-hour.

And

1KWh= 36,00,000 joules

= 3.6*10⁶ joules

14) How much energy is consumed when a current of 5 amperes flows through the filament (or element) of a heater having resistance of 100 ohms for two hours? Express it in joules.

Ans:

Given that,

I= 5A

R= 100 ohms

Time = 2 hrs = 2*60*60= 7200 seconds

We know that,

P= I²R

P= 25*100= 2500 W

Hence, energy consumed will be given by,

E= P*time

= 2500*7200= 18,000,000 J

E= 18*10⁶ joules

15) An electric bulb is connected to a 220 V power supply line. If the bulb draws a current of 0.5 A, calculate the power of the bulb.

Ans:

Given that,

V= 220V

I= 0.5A

We know that, an electric power is given by,

P= VI = 220*0.5= 110W

16) In which of the following cases more electrical energy is consumed per hour?

(i) A current of 1 ampere passed through a resistance of 300 ohms.

(ii) A current of 2 amperes passed through a resistance of 100 ohms.

Ans:

Given that,

I= 1A

R= 300 ohms

Time = 1 hr = 60*60= 3600 seconds

We know that,

P= I²R = 300 W

Energy consumed will be,

E= P*time = 300*3600= 1,080,000 J

Given that,

I= 2A

R= 100 ohms

We know that,

P= I²R = 4*100 = 400 W

Energy consumed will be given by,

E= P*time = 400*3600= 1,440,000 J

Thus, energy consumed will be more in b).

17) An electric kettle rated at 220 V, 2.2 kW, works for 3 hours. Find the energy consumed and the current drawn.

Ans:

Given that,

V= 220V

P= 2.2 KW

Time = 3hrs = 3*3600= 10800 seconds

We know that,

P= VI

I= P/V = 2.2*10³/ 220 = 0.01*103= 10 A

Thus, the current drawn will be 10A.

And the electrical energy consumed will be,

E= P*time

= 2.2*3= 6.6KWh

Thus, the energy consumed will be 6.6 KWh.

18) In a house two 60 W electric bulbs are lighted for 4 hours, and three 100 W bulbs for 5 hours every day. Calculate the electric energy consumed in 30 days.

Ans:

Given that,

P= 60 W

For 4 hours per day, then for 30 days: time = 120 hrs

Thus, the energy consumed by two 60W bulb in 30 days will be,

E= 2P*t = 2*60*120= 14,400 Wh

Now, three 100W bulbs glow for 5 hrs every day.

Hence, P= 100W

Time= 30*5= 150 hrs

Then, the energy consumed by three 100W bulbs is given by,

E= 3*P*time = 3*100*150= 45,000 Wh

Thus, the total energy consumed in 30 days will be,

E= 14,400 + 45,000= 59400 Wh = 59.4 KWh

19) A bulb is rated as 250 V; 0.4 A. Find its: (i) power, and (ii) resistance.

Ans:

Given that,

V= 250V

I= 0.4A

We know that, power is given by,

P= VI = 250*0.4= 100 W

And we know that,

V= IR

R= V/ I = 250/0.4= 625 ohm

20) For a heater rated at 4 kW and 220 V, calculate: (a) the current,

(b) the resistance of the heater,

(d) the cost if 1 kWh is priced at 4.60.

Ans:

P= 4kW

V= 220 V

We know that,

P= VI

I= P/V = 4*103/220= 0.01818*103= 18.18 A

We know that,

V= IR

R= V/I = 220/18.18= 12.1 ohms

Given that,

Time = 2 hrs

P= 4KW

E= P*time = 4*2= 8KWh

Given that, 1KWh is priced at 4.6 rs.

Then, total cost of energy consumed will be,

Total cost= 8*4.6= 36.8 rs

21) An electric motor takes 5 amperes current from a 220 volt supply line. Calculate the power of the motor and electrical energy consumed by it in 2 hours.

Ans:

Given that,

I= 5A

V= 220 V

Time = 2hrs= 7200 seconds

We know that,

The power of the motor is given by,

P= VI= 220*5= 1100 W= 1.1 KW

Electrical energy consumed will be given by,

E= P*time = 1100*2= 2200 Wh = 2.2 KWh

22) Which uses more energy: a 250 W TV set in 1 hour or a 1200 W toaster in 10 minutes?

Ans:

Given that,

P= 250W

Time = 1 hr

Energy consumed will be,

E= P*t

= 250*1

= 250 Wh

= 250*60*60

= 900 kW

Given that,

P= 1200 W

Time = 10min = 10*60= 600 seconds

Energy consumed will be,

E= P*t = 1200*600= 720,000 W = 720 kW

The energy consumed by TV is 900 KW and energy consumed by toaster is 720 kW.

So, TV uses more energy.

23) Calculate the power used in the 2 Ω resistor in each of the following circuits:

(i) a 6 V battery in series with 1 Ω and 2 Ω resistors.

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Ans:

Given that,

V= 6V

Total resistance= 1+2= 3 ohms

Now, current flowing through the circuit is,

V= IR

I= V/R

= 6/3

= 2 A

Thus, 2A current flows through each resistor.

Now, the voltage across the 2 ohm resistor will be,

V= IR

= 2*2

= 4V

Thus, the power used in 2 ohm resistor will be,

P= VI

= 4*2

= 8W

Given that,

V= 4V

Then, current flowing through 2 ohm resistor will be,

V= IR

I= V/R = 4/2= 2 A

Thus, the power used in 2 ohm resistor will be,

P= VI

= 4*2

= 8 W

24) Two lamps, one rated 40 W at 220 V and the other 60 W at 220 V, are connected in parallel to the electric supply at 220 V.

(a) Draw a circuit diagram to show the connections.

(b) Calculate the current drawn from the electric supply.

(c) Calculate the total energy consumed by the two lamps together when they operate for one hour.

Ans:

Following diagram shows the circuit diagram showing above connections.

Given that,

From figure,

P= VI

I₁= P₁/V = 40/220= 0.1818 A

And I₂= P₂/V

= 60/220= 0.2727 A

Thus, the total current drawn will be,

I= I₁ + I₂ = 0.1818 + 0.2727= 0.45A

c) Time = 1hr

Energy consumed by first lamp,

E₁= P₁*t= 40*1= 40 Wh

Energy consumed by second lamp,

E₂= P₂*t = 60*1= 60 Wh

Now, total energy consumed by both lamps together will be,

E = E₁ + E₂= 100 Wh = 0.1 kWh

We know that, 1kWh = 3.6*10⁶ joules

Hence,

0.1kWh= 0.1*3.6*10⁶= 3.6*10⁵ J = 360*10³ J = 360 kJ

28) An electric kettle connected to the 230 V mains supply draws a current of 10 A.

Calculate: (a) the power of the kettle. (b) the energy transferred in 1 minute.

Ans:

Given that,

V= 230V

I= 10A

The power of the kettle is given by,

P= VI= 230*10= 2300 W

Time = 1min = 60 seconds

The energy transferred will be,

E= P*t = 2300*60= 138,000 joules = 138 kJ

26) A 2 kW heater, a 200 W TV and three 100 W lamps are all switched on from 6 p.m. to 10 p.m. What is the total cost at Rs- 5.50 per kWh?

Ans:

Given that,

P₁= 2kW

P₂= 200 W

P₃= 100W

Time= 4 hrs

Total power will be given by,

P= P₁ + P₂+ 3P₃

= 2000 + 200 + 300

= 2500 W

Now total energy consumed will be,

E= P*t = 2500*4= 10,000 Wh = 10 KWh

Given that, the cost per KWh is 5.5 rs.

Then total cost will be,

Total cost= 10*5.5 = 55 rs

27) What is the maximum power in kilowatts of the appliance that can be connected safely to a 13 A; 230 V mains socket?

Ans:

Given that,

I= 13A

V= 230V

The maximum power in kilowatts of the appliance will be,

P= VI = 230*13= 2990W = 2.99 kW

28) An electric fan runs from the 230 V mains. The current flowing through it is 0.4 A. At what rate is electrical energy transferred by the fan?

Ans:

Given that,

V= 230V

I= 0.4A

P= VI= 230*0.4= 92 W

Thus, the electrical energy transferred by the fan will be 92 J/s.

Long Answer Type Question

29) (a) What is meant by “electric power”? Write the formula for electric power in terms of potential difference and current.

(b) The diagram below shows a circuit containing a lamp L, a voltmeter and an ammeter. The voltmeter reading is 3 V and the ammeter reading is 0.5 A.

(i) What is the resistance of the lamp?

(ii) What is the power of the lamp?

(c) Define kilowatt-hour. How many joules are there in one kilowatt-hour?

(d) Calculate the cost of operating a heater of 500 W for 20 hours at the rate of ` 3.90 per unit.

Ans:

Electric power is nothing but the electrical work done per unit time.

Power = world one/ time taken

P= W/t

Electric power is the rate at which electrical energy is consumed.

P= W/t

But, W= VIt

P= VIt/t = VI

Thus, P= V*I

Given that,

V= 3V

I= 0.5 A

The resistance of the lamp is given by,

V= IR

R= V/I = 3/0.5= 6 ohm

The power of the lamp is given by,

P= VI = 3*0.5= 1.5 W

One kilowatt hour is the amount of energy consumed when an electrical appliance has power rating of 1kilowatt which is used for 1 hour.

And commercial unit of electrical energy is the kilowatt hours.

1 Kilowatt hours= 36,00,000 joules

1KWh = 3.6*10⁶J

Given that,

P= 500W

Time = 20 hours

Energy consumed will be,

E= P*t = 500*20= 10,000 Wh = 10KWh

Given that, 3.9 rs is the cost for 1KWh.

Hence, the cost of operating heater will be,

Cost= 10*3.9= 39 rs.

Heating Effect of Current

Very Short Answer Type Questions

1) How does the heat H produced by a current passing through a fixed resistance wire depend on the magnitude of current I?

Ans:

The heat H produced by the current passing through a fixed resistance wire is directly proportional to the square of the magnitude of the current.

2) If the current passing through a conductor is doubled, what will be the change in heat produced?

Ans:

If the current passing through a conductor is doubled then the heat produced will be four times.

3) Name two effects produced by electric current.

Ans:

The two effects produced by electric current are heating effect and magnetic effect.

4) Which effect of current is utilised in an electric light bulb?

Ans:

Heating effect of an electric current is used in an electric light bulb.

5) Which effect of current is utilised in the working of an electric fuse?

Ans:

The heating effect of electric current is used in the working of an electric fuse.

6) Name two devices which work on the heating effect of electric current.

Ans:

Following are the two devices which works on the heating effect of electric current:
Light bulbs, electric fuse, heaters, electric iron etc.

7) Name two gases which are filled in filament type electric light bulbs.

Ans:

Argon and nitrogen are the two gases which are filled in filament type electric bulbs.

8) Explain why, filament type electric bulbs are not power efficient.

Ans:

As most of the power consumed by the filament of an electric bulb appears as heat due to which bulb becomes hot and only small amount of electric power is converted into light.
Hence, filament type electric bulbs are not power efficient.

9) Why does the connecting cord of an electric heater not glow hot while the heating element does?

Ans:

The heating element of an electric heater is made from nichrome which glows red hot because of large amount of heat is generated when current is passed through it. Since it has very high resistance.
But, the cord of an electric heater is made from copper which has very low resistance and hence heat produced will be negligible when current is passed through it.

Short Answer Type Questions

10) (a) Write down the formula for the heat produced when a current I is passed through a resistor R for time t.

(b) An electric iron of resistance 20 ohms draws a current of 5 amperes. Calculate the heat produced in 30 seconds.

Ans:

The formula for heat produced when a current I is passed through a resistor R for time t is given by,

H= I²*R*t

Given that,

R= 20 ohm

I= 5A

Time = 30 seconds

Heat generated will be given by,

H= I²*RT= 25*20*30= 15,000 J

11) State three factors on which the heat produced by an electric current depends. How does it depend on these factors?

Ans:

The three factors on which heat produced by the electric current depends are as follows:

Current through wire
Resistance of wire
Time for which current is passed through wire.

And the heat produced is directly proportional to the

Square of current I2 passing through wire
Resistance R is the wire
Time t for which current is passed through the wire.

Hence, H= I²*Rt

12) (a) State and explain Joule’s law of heating.

(b) A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination of resistances in half a minute.

Ans:

According to Joule’s law of heating, when an electric current is passed through the wire of high resistance such as nichrome wire then the wire becomes very hot and heat will be produced. This effect is called as Joule’s heating effect of electric current.

For example:

When an electric current is passed through the very thin and high resistance tungsten filament of an electric bulb then filament becomes white hot and emits light.
Tungsten metal has very high melting point due to which it is used in making filaments in electric bulbs.

Given that,

V= 220 V

Time = 30 seconds

Total resistance = 40+60= 100 ohm

But, V= IR

I= V/R= 220/100= 2.2A

Thus, heat produced will be,

H= I²*R*t = 2.2*2.2*100*30= 14520 J

13) Why is an electric light bulb not filled with air? Explain why argon or nitrogen is filled in an electric bulb.

Ans:

If the bulb is filled with air then the very hot tungsten filament get burnt in the oxygen of air.
Due to which electric bulbs are filled with chemically unreactive gases like nitrogen or argon or mixture of both.
When the tungsten filament is hot at that time these gases does not react with tungsten filament and hence the filament of electric bulb gets protected.
Hence, electric bulbs are filled with gases like nitrogen, argon or mixture of both not filled with air.

14) Explain why, tungsten is used for making the filaments of electric bulbs.

Ans:

Tungsten metal is used for making filaments of electric bulbs because it has very high melting point near about 3380°C.
Due to which when the tungsten filament becomes white hot then also it does not melt die to its high melting point.
Also, tungsten has high flexibility and low rate of evaporation at higher temperatures.
Because of all these reasons tungsten metal is used in making filaments of electric bulbs.

15) Explain why, the current that makes the heater element very hot, only slightly warms the connecting wires leading to the heater.

Ans:

Because the heating coils of all electrical appliances are made from an alloy nichrome which has high resistance and due to which when current is passed through them large amount of heat will be produced.
But the connecting wires are made from copper which has very low resistance and hence negligible heat will be produced in the connecting wires.

16) When a current of 4.0 A passes through a certain resistor for 10 minutes, 2.88 × 104 J of heat are produced. Calculate:

(a) the power of the resistor.

(b) the voltage across the resistor.

Ans:

Given that,

I= 4A

Time = 10 minutes = 600 seconds

H= 2.88*10⁴ J

Power of the resistor is given by,

P= I²R

But, H= I²Rt

2.88*10⁴ = 16*R*600

R= 2.88*10⁴/ 16*600 = 0.0003*10⁴ = 3 ohms

Now, P= I²R

= 16*3= 48 W

b) The voltage across the resistor will be,

P= VI

V= P/I = 48/4= 12 V

17) A heating coil has a resistance of 200 W. At what rate will heat be produced in it when a current of 2.5 A flows through it?

Ans:

Given that,

R= 200 ohm

I= 2.5A

Power is given by,

P= I²R = 6.25*200= 1250 W

Thus, the 1250 J/s is the rate at which heat is heat is developed in the heater.

18) An electric heater of resistance 8 W takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater.

Ans:

Given that,

R= 8 ohm

I= 15A

We know that,

P= I²R = 225*8= 1800 W

Thus, 1800 J/s is the rate at which heat is developed in the heater.

19) A resistance of 25 W is connected to a 12 V battery. Calculate the heat energy in joules generated per minute.

Ans:

Given that,

R= 25 ohm

V= 12V

We know that,

V= IR

I= V/R = 12/25= 0.48 A

Now,

Heat generated will be given by,

H= I²*R*t = 0.48*0.48*25*60

= 345.6 J

Thus,345.6 J is the energy generated per minute.

20) 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor?

Ans:

Given that,

H= 100J

R= 4 ohm

Time = 1 second

We know that,

H= I²*R*t

I2= H/R*t

I2= 100/4= 25

I = 5 A

We know that,

V= IR = 5*4= 20 V

Thus, the potential difference across the resistor will be 20V.

Long Answer Type Question

21) (a) Derive the expression for the heat produced due to a current ‘I’ flowing for a time interval ‘t’ through a resistor ‘R’ having a potential difference ‘V’ across its ends. With which name is this relation known?

(b) How much heat will an instrument of 12 W produce in one minute if it is connected to a battery of 12 V?

(c) The current passing through a room heater has been halved. What will happen to the heat? produced by it?

(d) What is meant by the heating effect of current? Give two applications of the heating effect of current.

(e) Name the material which is used for making the filaments of an electric bulb.

Ans:

Let us consider the current I is flowing through conductor of resistance R for time t, then work done by the current I when it flows through a resistor R for time t is given by,

W= Q*V

But, I= Q/t

Q= I*t

And V= IR

Thus, W= Q*V

W= I*t*I*R = I²*R*t

We assume that all the electrical work done or all the energy consumed will be converted into heat energy H and it is given by,

H= I²*R*t

This is the expression for heat produced in joules when a current of I ampere flows through the wire of R ohm resistance for the time of t seconds.

And this is the Joule’s law of heating.

Given that,

P= 12W

Time = 60 seconds

V= 12 V

We know that,

P= VI

I= P/V = 12/12= 1A

And by Ohm’s law,

V= IR

R= V/I = 12/1 = 12 ohm

Now, the heat produced will be given by,

H= I²*R*t = 1*12*60= 720 J

Given that,

I’= I/2

We know that,

H= I²*R*t

If I’= I/2 then heat produced will be,

H’= I’²*R*t

= I²*R*t/4

Thus, the heat produced will be one fourth.

When an electric current is passed through the high resistance wire such as nichrome alloy then it becomes very hot and produces heat, and this effect is called as heating effect of electric current.
And the heat produced will be, H= I²Rt
The following are the applications of heating effect of an electric current:
The heating effect is used to produce large amount of heat in electrical appliances such as electric iron, electric kettle, electric toaster, electric oven, room heater, water heaters etc.
Also, the heating effect is used in electric bulbs to produce light.

e) Tungsten metal with high melting point is used for making the filaments of an electric bulb.

Here is your solution of Lakhmir Singh Class 10 Physics 1th Chapter Electricity

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For more Chapter solution, click below

Lakhmir Singh Manjit Kaur Class 10 Physics 1st Chapter “Electricity” solution

Lakhmir Singh Class 10 Physics 1th Chapter Solution

(First Part)

Very short answer type questions:

Short answer type questions:-

Long answer type questions:

(Second Part)

Electric Current

Very Short Answer Type Questions-

Short Answer Type Questions

Long Answer Type Question

(Third Part)

Ohm’s Law

Very Short Answer Type Questions –

Short Answer Type Questions

Long Answer Type Question

Factors affecting the resistance of the conductor:

Domestic Electric Circuits: Series or Parallel

Electric Power

Heating Effect of Current

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