**Selina Concise Class 10 Physics Solution Chapter No. 8 – ‘Current Electricity’ For ICSE Board Students.**

Selina Concise Class 10 Physics Chapter 8 Current Electricity Exercise All Questions and Answers by Physics Teacher here in this post.

**Exercise 8 (A)**

**Question: 1**

**Define the term current and state its S.I. unit.**

**Solution:** The rate of flow charge is current. Ampere is the S.I. unit of current.

**Question: 2**

**Define the term electric potential. State it’s S.I. unit.**

**Solution:** The potential at a point is defined as the amount of work done per unit charge in bringing s positive test charge from infinity to the point. Volt is the S.I. unit of electric potential.

**Question: 3**

**How is the electric potential difference between the two points defined? State its S.I. unit.**

**Solution:** The potential difference between two point is equal to the work done per unit charge in moving a positive test charge from one point to the other.Joule per coulomb is the S.I unit potential difference.

**Question: 4**

**Explain the statement ‘the potential difference between two points is 1 volt’.**

**Solution:** The potential difference between two points is said to be 1 volt if the work done in moving 1 coulomb charge from 1 point to the other is 1 joule.

**Question: 5**

**(a.) State whether the current is a scalar or vector? What does the direction of current convey?**

**(b.) State whether the potential is a scalar or vector? What does the positive and negative sign of potential convey?**

**Solution:**

a.) Current is the scalar quantity. The direction of flow of current is opposite to the direction of electrons.

b.) Potential is the scalar quantity. The work done due to attractive force is negative and work done due to repulsive force is positive.

**Question: 6**

**Define the term resistance. State its S.I. unit.**

**Solution:** The obstruction offered to the flow of current by the conductor is called its resistance. Ohm is the S.I. unit of resistance.

**Question: 7**

**(a) Name the particles which are responsible for the flow of current in a metallic wire.**

**(b) Explain the flow of current in a metallic wire on the basis of movement of the particles named by you above in part (a).**

**(c) What is the cause of resistance offered by the metallic wire in the flow of current through it?**

**Solution:**

a.) Free electrons in the metallic wire are responsible for the flow of current.

b.) The free electrons available in the metallic wire will carry current. N number of electrons pass through time t.

Charge = N x charge on electrons.

c.) A metal is electrically neutral. Hence it contains both positive and negative charge. The flow of free electrons is repelled by the positive charge in the metal. Hence, the resistance is offered by the metallic wire in the flow of current through.

**Question: 8**

**State Ohm’s law and draw a neat labelled circuit diagram containing a battery, a key, a voltmeter, an ammeter, a rheostat and an unknown resistance to verify it.**

**Solutions: **The current flowing in a conductor is directly proportional to the potential difference applied across its ends provided that the physical conditions and the temperature of the conductor remain constant, this is ohm’s law.

**Question: 9**

**(a.) Name and state the law which relates the potential difference and current in a conductor.**

**(b.) What is the necessary condition for a conductor to obey the law named above in part (a)?**

**Solution:**

a.) The current flowing in a conductor is directly proportional to the potential difference applied across its ends provided that the physical conditions and the temperature of the conductor remain constant, this is ohm’s law.

b.) The condition for the ohm’s law is that the temperature of the conductor shouldremain constant.

**Question: 10**

**(a.) Draw a V-I graph for a conductor obeying Ohm’s law.**

**(b.) What does the slope of V-I graph for a conductor represent?**

**Solution:** (a) voltage to current graph for conductor obeying ohm’s law.

(b.) Slope represents the resistance.

**Question: 11**

**Draw a I-V graph for a linear resistor. What does its slope represent?**

**Solution: **current voltage graph for a linear resistor.

**Question: 12**

**What is an ohmic resistor? Give one example of an ohmic resistor. Draw a graph to show its current – voltage relationship. How is the resistance of the resistor determined from this graph?**

**Solution: **the conductor behaving according ohm’s law is known as ohmic resistors.

The resistance can be determined by the slope of above graph.

**Question: 13**

**What are non-ohmic resistors? Give one example and draw a graph to show its current-voltage relationship.**

**Solution:** The conductor which do not obey ohm’s law is known as non-ohmic resistors

**Question: 14**

**Give two differences between an ohmic and non-ohmic resistor**

**Solution:**

Ohmic resistor |
Non-ohmic resistor |

It obeys ohm’s law. | It does not obey ohm’s law. |

The graph for the potential difference V versus current I is a straight line passing through the origin. | The graph for the potential difference V versus current I is not a straight line, but a curve which may pass through the origin. |

**Question: 15**

**Fig. below shows the I-V curves for two resistors. Identify the ohmic and non-ohmic resistors. Give a reason for your answer.**

**Solution:** Graph (a) shows the non-ohmic resistor and graph (b) is ohmic resistor. The graph (a) shows curve which is characteristic of non-ohmic resistor and graph (b) shows the straight line which is characteristic of ohmic resistor.

**Question: 16**

**Draw a V – I graph for a conductor at two different temperatures. What conclusion do you draw from your graph for the variation of resistance of conductor with temperature?**

**Solution:**

Based on the graph the temperature of the wire A is more than the temperature of wire B as the straight line of A is steeper than line of B. hence, the wire A offers more resistance than wire B hence wire A is thinner as the thinner wire more resistance to raise the temperature of the wire.

**Question: 17**

**(a.) How does the resistance of a wire depend on its radius? Explain your answer.**

**(b.) Two copper wires are of same length, but one is thicker than the other. Which will have more resistance?**

**Solution:**

a.) The resistance of wire is inversely proportional to the radius of the wire. The thin wire offers more resistance than thin wire.

b.) The copper wire with thin cross section will offer more resistance. As they have same length.

**Question: 18**

**How does the resistance of a wire depend on its length? Give a reason of your answer.**

**Solution:** The resistance of a wire is directly proportional length of wire. The resistance offered by the wire is depends upon the repulsion between differently charged particles. Hence, the more the length more the resistance offered by the wire as the electrons have to travel more distance under influence of repulsion.

**Question: 19**

**How does the resistance of a metallic wire depend on its temperature? Explain with reason.**

**Solution:** The resistance of wire increases as the temperature increases. The increase in the temperature causes more random motion of electrons which ultimately increases the resistance of the wire.

**Question: 20**

**Two wires, one of copper and other of iron, are of the same length and same radius. Which will have more resistance? Give reason.**

**Solution:** Resistivity of the material is directly proportional to the resistance offered by that material. The resistivity of iron is more iron is more than that of copper. hence, the resistance of iron is more than wire of copper as the same length and radius is same.

**Question: 21**

**Name three factors on which the resistance of a given wire depends and state how it is affected by the factors stated by you?**

**Solution:**

a.) The length of wire. The resistance of wire is directly proportional to the length of wire. More the length of wire more the resistance.

b.) Cross section of wire. The resistance of wire is inversely proportional to the cross section of wire. Thinner the wire more the resistance offered by the wire.

**Question: 22**

**Define the term specific resistance and state its S.I. unit.**

**Solution:**The resistance of a wire of a material of unit length and unit area of cross section. Ohm x meter is the S.I. unit of specific resistance.

**Question: 23**

**Write an expression connecting the resistance of a wire and specific resistance of its material. State the meaning of symbols used.**

**Solution:**

Specific resistance (ρ) = R a / l

Where,

R = resistance of wire

a = area of cross section

l = length of wire.

**Question: 24**

**State the order of specific resistance of (i) a metal, (ii) a semiconductor and (iii) an insulator.**

**Solution:**

i.) The specific resistance of metal is low as it has high conductivity.

ii.) The specific resistance of semi-conductor is more than metal

iii.) The specific resistance of insulator is highest, as it doesn’t show any conductivity.

**Question: 25**

**(a.) Name two factors on which the specific resistance of a wire depends?**

**(b.) Two wires A and B are made of copper. The wire A is long and thin while the wire B is Short and thick. Which will have more specific resistance?**

**Solution:**

a.) Material and temperature of the substance are the two factors specific resistance of a wire depends upon.

b.) The specific resistance is same as the specific resistance is depends upon the material of the wire.

**Question: 26**

**Name a substance of which the specific resistance remains almost unchanged by the increase in temperature.**

**Solution:** Manganin is the substance of which the specific resistance almost remains unchanged by the increase in the temperature.

**Question: 27**

**How does specific resistance of a semi-conductor change with the increase in temperature?**

**Solution:** The specific resistance of semi-conductors decreases with increase in the temperature.

**Question: 28**

**How does (a) resistance, and (b) specific resistance of a wire depend on its (i) length, and (ii) radius?**

**Solution:**

a.) Resistance is directly proportional to the length of wire and inversely proportional to the cross section.

b.) Specific resistance is independent of length and cross section of the wire.

**Question: 29**

**(a) Name the material used for making the connection wires. Give reason for your answer.**

**(b) Why should a connection wire be thick?**

**Solution:**

a.) Copper or aluminium are used in making of wire as the they have low specific resistance. Hence, there wire offers very low resistance.

b.) As the resistance of a wire is inversely proportional to the cross section of the wire. Connections are made thick as the resistance decreases with it.

**Question: 30**

**Name the material used for making a fuse wire. Give a reason.**

**Solution:** Alloy of tin and lead is used to make fuse wire as it has high resistivity and low melting point. Which is important for working of the fuse.

**MULTIPLE CHOICE TYPE**

**Question: 1**

**Which of the following is an ohmic resistance?**

(a) LED

(b) Junction diode

(c) Filament of a bulb

(d) Nichrome wire

**Solution:** (d)Nichrome wire

**Question: 2**

**For which of the following substances, resistance decreases with increase in temperature?**

(a) Copper

(b) Mercury

(c) Carbon

(d) Platinum

**Solution:** (c) Carbon

**NUMERICALS**

**Question: 1**

**In a conductor, 6.25 × 10 ^{16} electrons flow from its end A to B in 2 s. Find the current flowing through the conductor. (e = 1.6 × 10-19 C)**

**Solution:** Given

Number of electrons, n = 6.25 x 10^{16}

Charge, e = 1.6 x 10^{-16}

Time, t = 2s

Current = ne / t

I = 6.25 x 1016 x 1.6 x 10-16 / 2

I = 5 x 10-3 A = 5 mA.

**Question: 2**

**A current of 1.6 mA flows through a conductor. If charge of an electron is ****-1.6 x 10 ^{-19 }**

**coulomb, find the number of electrons that will pass each second through the cross section of that conductor.**

**Solution:** Given

Current, I = 1.6 mA = 1.6 x 10^{-3} A

Charge, e = 1.6 x 10^{-16}C

Time = 1s

number of electrons = I x t / e

n= 1.6 x 10^{-3} x 1 / 1.6 x 10^{-16}

n = 10^{16}

**Question: 3**

**Find the potential difference required to flow a current of 200 mA in a wire of resistance 20 ohm.**

**Solution:** Given

Current, I = 200 mA = 0.2 A

Resistance, R = 20 ohm

Potential difference, V = I x R

V = 0.2 x 20

V = 4 V

**Question: 4**

**An electric bulb draws 1.2 A current at 6.0 V. Find the resistance of filament of bulb while glowing.**

**Solution:** Given

Current, I = 1.2 A

Potential difference, V = 6 V

Resistance, R = V / I

R = 6 / 1.2

R = 5 ohm.

**Question: 5**

**A car bulb connected to a 12 volt battery draws 2 A current when glowing. What is the resistance of the filament of the bulb? Will the resistance be more, same or less when the bulb is not glowing.**

**Solution:** Given

Potential difference, V = 12 V

Current, I = 2 A

Resistance of filament, R = V / I

R = 12 / 2

R = 6 ohm

When the bulb is not glowing, the resistance will be less.

**Question: 6**

**Calculate the current flowing through a wire of resistance 5 Ohm connected to a battery of potential difference 3 V.**

**Solution:** Given

Resistance, R = 5 ohm

Potential difference, V = 3V

Current, I = V / R

I = 3 / 5

I = 0.6 A

**Question: 7**

**In an experiment of verification of Ohm’s law, following observations are obtained.**

Potential difference V (in volt) | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 |

Current I (in amp) | 0.2 | 0.4 | 0.6 | 0.8 | 1.0 |

Draw a characteristic V-I graph and use this graph to find:

(a.) potential difference V when the current I is 0.5 A.

(b.) current I when the potential difference V is 0.75 V.

(c.) resistance in circuit

Solution: (a)

Potential different is 1.25V when current is 0.5 A.

(b) when potential difference is 0.75 V current is 0.3 A.

(c) resistance = voltage / current

R = 1.25 / 0.5

R = 2.5 ohm

**Question: 8**

**Two wires of the same material and same length have radii 1 mm and 2 mm respectively. Compare (i) their resistances (ii) their specific resistance.**

**Solution:**

i.) The resistance

R1 = ρ (I / πr_{1}^{2})

R2 = ρ (I / πr_{2}^{2})

R1 / R2 = r_{2}^{2 }/ r_{1}^{2}

R1 / R2 = 4/1

Ratio of resistance = 4:1

ii.) The ratio of specific resistance will be 1:1 as the materials is same.

**Question: 9**

**A given wire of resistance 1 Ohm is stretched to double its length. What will be its new resistance?**

**Solution:**

Resistance, R1 = ρ (I / πr^{2})

Now, length of wire

Resistance, R2 =ρ (2I /πr_{1}^{2}/2)

R2 = 4 ρ (I / πr^{2})

Therefore,

R2 = 4 x R1

R2 = 4 ohm.

**Question: 10**

**A wire of resistance 3 Ohm and length 10 cm is stretched to length 30 cm. Assuming that it has a uniform cross-section, what will be its new resistance?**

**Solution:** Given

Resistance, R = 3 ohm

Length, l1 = 10cm

Length, l2 = 30 cm = 3l

Area will be πr^{2}/3

New resistance = ρ (I / πr^{2})

R2 = ρ (3I / πr^{2}/3)

R2 = 9 ρ (I / πr^{2})

R2 = 9 R

R2 = 9 x 3

R2 = 27 ohm.

**Question: 11**

**A wire of resistance 9 Ohm having length 30 cm is tripled on itself. What is its new resistance?**

**Solution:** Given

Resistance = 9 ohm

Length = 30 cm

New length = 1/3L

Area will be 3πr^{2}

New resistance = ρ (I / πr^{2})

R2 = ρ (I/3 / 3πr^{2})

R2 = 1/9 ρ (I / πr^{2})

R2 = 1/9 R

R2 = 9 /9

R2 = 1 ohm.

**Question: 12**

**What length of copper wire of specific resistance 1.7 x 10 ^{-8} ohmand radius 1 mm is required so that its resistance is 1 ohm.**

**Solution:** Given

specific resistance = 1.7 x 10^{-8 }ohm

radius = 1mm = 10^{-3} m

resistance = ρ (I / πr^{2})

l = R πr^{2}/ ρ

l = 1 x 3.14 x 10^{-6}/ 1.7 x 10^{-8}

l = 184.7 m

**Question: 13**

**The filament of a bulb takes a current 100 mA when potential difference across it is 0.2 V. When the potential difference across it becomes 1.0 V, the current becomes 400 mA. Calculate the resistance of filament in each case and account for the difference.**

**Solution:** Given

Current, I = 100 mA = 0.1 A

Potential difference, V = 0.2 V

Resistance = V / I

= 0.2 / 0.1

= 2 ohm

Current, I = 400 mA = 0.4 A

Potential difference, V = 1 V

Resistance = V / I

= 1 / 0.4

= 2.5 ohm

**Exercise 8 (B)**

**Question: 1**

**Explain the meaning of the terms e.m.f., terminal voltage and internal resistance of cell.**

**Solution: **The e.m.f. of a cell defined as the energy per unit charge in to positive test charge around the complete of the cell.

The work done per unit charge in carrying a positive test charge around the circuit connected across the terminal of the cell is called as terminal voltage.

The resistance offered by the electrolyte inside the cell, to the flow of current is called as the internal resistance of the cell.

**Question: 2**

**State two differences between the e.m.f. and terminal voltage of a cell.**

**Solution:**

e.m.f. of the cell. |
Terminal voltage of the cell. |

The energy per unit charge in to positive test charge around the complete of the cell. | The work done per unit charge in carrying a positive test charge around the circuit connected across the terminal of the cell |

It is characteristic of an cell. | It depends upon the amount of current drawn from the cell. |

**Question: 3**

**Name two factors on which the internal resistance of a cell depends and state how does it depend on the factors stated by you.**

**Solution:**

The surface area of the electrodes will affect the internal resistance as the surface area of the electrode increases the internal resistance of the cell decreases. The internal resistance is directly proportional to the distance between the electrodes.

**Question: 4**

**A cell of e.m.f. ε and internal resistance r is used to send current to an external resistance R. Write expressions for (a) the total resistance of circuit, (b) the current drawn from the cell, (c) the p.d. across the cell, and (d) voltage drop inside the cell.**

**Solution:**

a.) Total resistance = R + r

b.) Current drawn =

e.m.f.ε = V + v

ε = IR + Ir

ε = I (R + r)

I = (R + r)

C.) Potential difference across the cell = (ε/ R+ r) / R

d.) Voltage drop inside the cell = (ε/ R+ r) / r

**Question: 5**

**A cell is used to send current to an external circuit. (a) How does the voltage across its terminal compare with its emf? (b) Under what condition is the emf of the cell equal to its terminal voltage?**

**Solution:**

a.) The terminal voltage V is less than e.m.f. of the cell as the current is drawn from a cell.

b.) The emf of the cell becomes equal to its terminal voltage when there is no current drawn.

**Question: 6**

**Explain why the p.d. across the terminals of a cell is more in an open circuit and reduced in a closed circuit.**

**Solution:** When the circuit is closed current flows through the circuit. Which causes the drop in the potential difference across the internal resistance of the cell. Hence, potential difference across the terminals of a cell is more in an open circuit and reduced in a closed circuit.

**Question: 7**

**Write the expressions for the equivalent resistance R of three resistors R1, R2 and R3 joined in (a) parallel, (b) series.**

**Solution:**

a.) Total resistance in parallel

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

b.) Total resistance in the series

R = R1 + R2 + R3

**Question: 8**

**How would you connect two resistors in series? Draw a diagram. Calculate the total equivalent resistance.**

**Solution:**

We can apply ohms law on the resistors separately

V1 = IR1

V2 = IR2

V = V1 + V1

IR = IR1 + IR2

R = R1 + R2

**Question: 9**

**Show by a diagram how two resistors ****R _{1} and R_{2 }**

**are joined in parallel. Obtain an expression for the total resistance of the combination.**

**Solution:**

We can apply ohms law on the resistors separately

V= I1R1

I1 = V / R1

V = I2R2

I2 = V / R2

I = I1 + I1

V / R = V / R1 + V / R2

1 / R = 1 / R1 + 1 / R2

**Question: 10**

**State how are the two resistors joined with a battery in each of the following cases when:**

**(a.) same current flows in each resistor**

**(b.) potential difference is same across each resistor.**

**(c.) equivalent resistance is less than either of the two resistances.**

**(d.) equivalent resistance is more than either of the two resistances.**

**Solution:**

(a.) When two resistors are connected in series.

(b.) When two resistors are connected in parallel.

(c.) When two resistors are connected in series.

(d.) When two resistors are connected in parallel.

**Question: 11**

**The V-I graph for a series combination and for a parallel combination of two resistors is shown in fig. Which of the two, A or B, represents the parallel combination? Give a reason for your answer.**

**Solution:**

The line B represent more change in voltage which is a result of more resistance. On the other hand line A represents lower change in voltage which accounts to lower resistance. When resistors are connected in parallel they show less resistance than when connected in series. Hence, line A represents parallel combination.

**MULTIPLE CHOICE TYPE**

**Question: 1**

**In series combination of resistances:**

(a.) P.d. is same across each resistance

(b.) Total resistance is reduced

(c.) Current is same in each resistance

(d.) All of the above are true

**Solution:** (c) Current is same in each resistance

**Question: 2**

**In parallel combination of resistances:**

(a.) P.D. is same across each resistance

(b.) Total resistance is increased

(c.) Current is same in each resistance

(d.) All of the above are true

**Solution:** (a) P.D. is same across each resistance

**Question: 3**

**Which of the following combinations have the same equivalent resistance between X and Y?**

**Solution:**

a.) Resistances are connected in parallel

1/R = ½ + ½ =1 ohm

b.) Resistances are connected in parallel

1/R = 1/1 + 1/1 +1/2 = 2.5 ohm

c.) Resistances are connected in parallel

1/R = 1/1 + 1/1 = 0.5 ohm

d.) Pair of resistors are connected in parallel

1/R = ½ + ½ = 1 ohm

Therefore, the (a) and (d) combination has same resistance across between X and Y.

**NUMERICALS**

**Question: 1**

**The diagram in figure shows a cell of e.m.f. ε = 2 volt and internal resistance r = 1 ohm connected to an external resistance R = 4 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammetere and voltmeter when (i) the key K is open, and (ii) the key K is closed**

**Solution:** Given

Emf, ε = 2 volt

Internal resistance, r = 1 ohm

External resistance, R = 4 ohm

i.) Voltage = ε – Ir

As the key is open I = 0

Voltage = 2 – 0 x 1

Voltage = 2 volt

ii.) Ammeter reading

I = ε / (R + r)

I = 2 / (4 + 1)

I = 2/5

I = 0.4 A

Voltage = ε – Ir

Voltage = 2 – 0.4 x 1

Voltage = 1.6 V

**Question: 2**

**A battery of e.m.f. 3.0 V supplies current through a circuit in which resistance can be changed. A high resistance voltmeter is connected across the battery. When the current is 1.5 A, the voltmeter reads 2.7 V. Find the internal resistance of the battery.**

**Solution:** Given

Emf, ε = 3 V

Current, I = 1.5 A

Voltage = 2.7 V

V = ε – Ir

2.7 = 3 – 1.5r

0.3 / 1.5 = r

r = 0.2 ohm.

**Question: 3**

**A cell of emf 1.8 V and internal resistance 2 ohm is connected in series with an ammeter of resistance 0.7 ohm and resistance of 4.5 ohm as shown in figure.**

**(a) What would be the reading of the ammeter?**

**(b) What is the potential difference across the terminals of the cell?**

**Solution:**

Given

Emf, ε = 1.8V

Internal resistance, r = 2 ohm

Ammeter = 0.7 ohm

External resistance, R = 4.5 ohm

Total resistance = 2 + 0.7 + 4.5

Total resistance = 7.2 ohm

a.) Reading of ammeter

Current = ε / R

I = 1.8 / 7.2

I = 0.25 A

b.) Total resistance excluding internal resistance = 4.5 + 0.7 = 5.2 ohm

Voltage = IR

V = 0.25 x 5.2

V = 1.3 V

**Question: 5**

**A cell of e.m.f. ε and internal resistance r sends current 1.0 A when it is connected to an external resistance 1.9 ohm. But its sends current 0.5 A when it is connected to an external resistance of 3.9 ohm. Calculate the values of e and r.**

**Solution:** Given

Emf, ε = 15 V

External resistance, R1 = 1.9 ohm

Current, I1 = 1 A

Eternal resistance, R2 = 3.9 ohm

Current, I2 = 0.3 A

For first case, ε = I (R + r)

= 1 (1.9 + r)

For second case, ε = I (R + r)

= 0.5 (3.9 +r)

Emf of cell is same. Therefore,

1 (1.9 + r) = 0.5 (3.9 + r)

r = 0.05 / 0.5

r = 0.1 ohm

emf, ε = 1.9 + r

ε = 1.9 + 0.1

ε = 2 V

**Question: 6**

**Two resistors having resistance 4 ohm and 6 ohm are connected in parallel. Find their equivalent resistance.**

**Solution:** Given

R1 = 4 ohm

R2 = 6 ohm

As they are connected in parallel

1/R = 1/ R1 + 1/ R2

1/R = 1/ 4 + 1/6

1/R = 10/24

R = 2.4 ohm

**Question: 7**

**Four resistors each of resistance 2 ohm are connected in parallel. What is the effective resistance?**

**Solution: **Given

R1 = 2 ohm

R2 = 2 ohm

R3 = 2 ohm

R4 = 2 ohm

As they are connected in parallel

1/R = 1/ R1 + 1/ R2 + 1/R3 + 1/R4

1/R = ½ + ½ + ½ + ½

1/R = 2

R = 0.5 ohm.

**Question: 8**

**You have three resistors of values 2 Ω, 3 Ω and 5 Ω. How will you join them so that the total resistance is less than 1 Ω? Draw diagram and find the total resistance.**

**Solution:** We have to connect the resistors in parallel.

Total resistance, R

1 / R = 1 / 2 + 1 / 3 + 1 / 5

1 / R = 31 / 30

R = 0.97 Ohm

**Question: 9**

**Three resistors each of 2 Ω are connected together so that their total resistance is 3 Ω. Draw a diagram to show this arrangement and check it by calculation.**

**Solution:** to get total resistance of 3 ohm we have to connect two resistors in parallel and one in series as shown below.

For resistors in parallel

1 / r = 1/2 + ½

r = 1ohm

now this is connected to 2 ohm resistor in series

total resistance = 1 + 2

R = 3 ohm.

**Question: 10**

**Calculate the equivalent resistance between the points A and B in figure if each resistance is 2.0 Ω**

**Solution:** Given

Resistor value = 2 ohm

The total resistance in parallel, 1/R = ½ + ½ = 1ohm

Therefore, total resistance of the circuit = 2 + 2 + 1

= 5 ohm

**Question: 11**

**A combination consists of three resistors in series. Four similar sets are connected in parallel. If the resistance of each resistor is 2 ohm, find the resistance of the combination.**

**Solution:** Given

Resistor value = 2 ohm

Four sets of three of these resistors are connected in series

Resistance of each one = 2 + 2 + 2 = 6 ohm

The resistance of all four set is 6 ohm.

Now these four sets are connected in parallel.

1/R = 1/6 + 1/6 + 1/6+ 1/6

1/R = 4/6

R = 6/4

R = 1.5 ohm.

**Question: 12**

**In the circuit shown below in figure, calculate the value of x if the equivalent resistance between the points A and B is 4 ohm.**

**Solution:** Given

R = 4 ohm

R1 = 4 ohm

R2 = 8 ohm

R3 = x ohm

R4 = 5 ohm

Now R1 and R2 are connected in series and R3 and R4 are connected in series

R1 + R2 = 4+ 8 = 12 ohm

R3 + R4 = x + 5 ohm

Now these are connected in parallel,

1/R = 1/R1+ R2 + 1/ R3+R4

¼ = 1/12 + 1/ x + 5

¼ – 1/12 = 1/ x + 5

2/12 = 1/ x + 5

1/6 = 1 / x + 5

x = 6 – 5

x = 1 ohm.

**Question: 13**

**Calculate the effective resistance between the points A and B in the circuit shown in figure**

**Solution:** The resistance in series in upper area = 1 +1 + 1 = 3 ohm

The resistance in series in lower area = 2 + 2 + 2 = 6 ohm

The total resistance in the parallel

1/R = 1/3 + ½ + 1/6

1/R = 2 + 3 + 1/ 6

1/R = 1 ohm

Total resistance across A to B = 1 + 1 + 1 = 3 ohm.

**Question: 14**

**A uniform wire with a resistance of 27 ohm is divided into three equal pieces and then they are joined in parallel. Find the equivalent resistance of the parallel combination.**

**Solution:** As the resistance was split in three pieces. Therefore, the resulting resistance

27/3 = 9 ohm

As they are connected in parallel

1/R = 1/9 + 1/9 + 1/9

1/R = 3/9

1/R = 1/3

R = 3 ohm

**Question: 15**

**A circuit consists of a 1 ohm resistor in series with a parallel arrangement of 6 ohm and 3 ohm resistors. Calculate the total resistance of the circuit. Draw a diagram of the arrangement.**

**Solution:** As the resistance is connected in parallel

1/R = 1/6 + 1/3

1/R = 3/6

R = 2 ohm

Total resistance = 2 + 1

= 3 ohm

**Question: 16**

**Calculate the effective resistance between the points A and B in the network shown below in figure.**

**Solution:** The parallel resistance

1/R = 1/12 + 1/6 + ¼

1/R = 6 /12

R = 2 ohm

Total resistance across A to B = 2 + 2 + 5

= 9 ohm

**Question: 17**

**Calculate the equivalent resistance between the points A and B in figure**

**Solution:** The resistance connected in series = 3 +2 = 5 ohm

The resistance connected in series in series = 6 + 4 = 10 ohm

Therefore, total resistance across A to B

1/R = 1/5 + 1/ 30 + 1/10

1/R = 6+ 1 + 3 /30

1/R = 1/3

R = 3 ohm

**Question: 18**

**In the network shown in adjacent figure, calculate the equivalent resistance between the points (a) A and B (b) A and C**

**Solution:**

a.) Resistance between A and B.

The resistance connected series = 2 + 2 +2 = 6 ohm.

The resistance connected in parallel

1/R = 1/6 + ½

1/R = 4/6

R = 1.5 ohm

b.) The resistance between A and C

Resistance in series = 2 + 2 = 4 ohm

Two of the 4 ohm resistors are connected in parallel.

1/R = ¼ + ¼

1/R = ½

R = 2 ohm

**Question: 19**

**Five resistors, each of 3 ohm, are connected as shown in figure. Calculate the resistance (a) between the points P and Q, and (b) between the points X and Y.**

**Solution:**

a.) The resistance between P and Q.

The resistance in series = 3 + 3 = 6 ohm

The total resistance between P and Q

1/R = 1/6 + 1/3

1/R = 3/6

R = 2 ohm

b.) Total resistance across X and Y

All the resistors connected in series

Total resistance = 3 + 2 + 3

Total resistance = 8 ohm.

**Question: 20**

**Two resistors of 2 ohm and 3 ohm are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.**

**Solution:**

**a.) In series**

Total resistance of the circuit, R

R = 2 + 3 = 5 ohm

V = 6 v

Current = V / R

I = 6 / 5 = 1.2 A

**b.) In parallel**

Total resistance of the circuit, R

1/R = ½ + 1/3

1/R = 5 / 6

R = 1.2 ohm

V = 6 V

Current, I = V / R

I = 6 / 1.2

I = 5 A

**Question: 21**

**A resistor of 6 ohm is connected in series with another resistor of 4 ohm. A potential difference of 20 V is applied across the combination. (a) Calculate the current in the circuit and (b) potential difference across the 6 ohm resistor.**

**Solution:**

**a.) As the resistors connected in series **

R = 6 + 4 = 10 ohm

*Current I = V / R *

I = 20 / 10

I = 2 A

**b.) To calculate potential difference across 6 ohm resistor **

R = 6 ohm

I = 2 A

V = IR

V = 6 x 2

V = 12 V

**Question: 22**

**Two resistors of resistance 4 Ω and 6 Ω are connected in parallel to a cell to draw 0.5 A current from the cell.**

**(a) Draw a labelled diagram of the arrangement**

**(b) Calculate current in each resistor.**

**Solution:**

a.)

**b.) Total resistance of the circuit **

1 / R = ¼ + 1/6

1/R = 10 /24

R = 2.4 ohm

I = 0.5 A

Voltage, V = IR

V = 0.5 x 2.4

V = 1.2

Therefore,

Current through resistor of 4 ohm

I = 1.2 / 4

I = 0.3 A

Current through resistor of 6 ohm

I = 1.2 / 6

I = 0.2 A

**Question: 23**

**Calculate the current flowing through each of the resistors A and B in the circuit shown in figure?**

**Solution:** Given

For resistor A = 1 ohm

Voltage = 2 V

Current across A,

I = V / R

I = 2 / 1

I = 2 A

For resistor B

Resistor = 2 ohm

I = V / R

I = 2 / 2

I = 1

**Question: 24**

**In figure, calculate:**

**(a.) the total resistance of the circuit.**

**(b.) the value of R, and**

**(c.) the current flowing in R**

**Solution:**

**a.) The total resistance of the circuit **

Given

Current, I = 0.4 A

Potential difference, V = 4 V

R1 = V / I

R1 = 4 / 0.4

R1 = 10 ohm.

**b.) To calculate value of R **

The total resistance is 10 ohm

As the two resistors are connected in parallel

1/10 = 1/R + 1/ 20

1/R = 1/20

R = 20 ohm

**c.) To calculate current flowing through R **

R = 20 ohm

V = 4 V

I = V / R

I = 4 / 20

I = 0.2 A

**Question: 25**

**A particular resistance wire has a resistance of 3.0 ohm per meter. Find:**

**(a.) The total resistance of three lengths of this wire each 1.5 m long, in parallel.**

**(b.) The potential difference of the battery which gives a current of 2 A in each of the 1.5 m length when connected in the parallel to the battery (assume that resistance of the battery is negligible).**

**(c.) The resistance of 5 m length of a wire of the same material, but with twice the area of cross section.**

**Solution:**

a.) Resistance of wire per metre is 3 ohm. Hence for 1.5m wire resistance is 4.5 ohm

Total resistance in parallel

1/R = 1/4.5 + 1/4.5 + 1/4.5

1/R = 3/4.5

R = 1.5 ohm

b.) Current, I = 2 A

Potential difference, V = IR

V = 2 x 4.5

V = 9 V

c.) Resistance of 1m wire is 3 ohm.

Therefore, 5 m length of wire resistance = 3 x 5 = 15 ohm

But as the cross section is doubled resistance is halved as they are inversely proportional.

Resistance = 15 /2 = 7.5 ohm.

**Question: 26**

**A cell supplies a current of 1.2 A through two resistors each of 2 ohm connected in parallel. When resistors are connected in series, it supplies a current of 0.4 A. Calculate: (i) the internal resistance and (ii) e.m.f. of the cell.**

**Solution:** Given

Current, I = 1.2 A

Resistance as in parallel

1/R = ½ + ½ = 1

R = 1 ohm

ε = I (R + r)

ε = 1.2 (1 + r)

resistance as in series,

R = 2 + 2 = 4 ohm

ε = 0.4 (4 + r)

as the emf is same,

1.2 (1 + r) = 0.4 (4 + r)

1.2r – 0.4r = 1.6 – 1.2

0.8r = 0.4

r = 0.5 ohm

emf of cell, ε = I (R + r)

ε = 1.2 (1 + 0.5)

= 1.2 + 0.6

= 1.8 V

**Question: 27**

**A battery of emf 15 V and internal resistance 3 ohm is connected to two resistors 3 ohm and 6 ohm connected in parallel. Find (a) the current through the battery (b) p.d. between the terminals of the battery (c) the current in 3 ohm resistor (d) the current in 6 ohm resistor.**

**Solution:** Given

Emf = 15 V

Internal resistance, r = 3 ohm

External resistance, 1/R = 1/3 + 1/6

1/R = 3/6

1/R = ½

R = 2 ohm

**a.) Current through the battery **

ε = I (R + r)

15 v = I (2 + 3)

I = 15 / 5

I = 3 A

**b.) Potential difference **

V = I R

V = 3 x 2

V = 6 V

**c.) Current across 3 ohm resistor **

I = V / R

I = 6 / 3

I = 2 A

**d.) Current across 6 ohm resistor**

I = V /R

I = 6 / 6

I = 1 A

**Question: 28**

**The circuit diagram in figure shows three resistors 2 ohm, 4 ohm and R ohm connected to a battery of e.m.f. 2 V and internal resistance 3 ohm. If main current of 0.25 A flows through the circuit, find:**

**(a.) the p.d. across the 4 ohm resistor**

**(b.) the p.d. across the internal resistance of the cell,**

**(c.) the p.d. across the R ohm or 2 ohm resistor, and**

**(d.) the value of R.**

**Solution:**

a.) Potential difference across 4 ohm resistor.

Resistance, R = 4 ohm

Current, I = 0.25 A

Potential difference, V= IR

V = 0.25 x 4

V = 1 V

b.) Potential difference across cell.

Internal Resistance, r= 3 ohm

Current, I = 0.25 A

Potential difference, V= Ir

V = 0.25 x 3

V = 0.75 V

c.) We have first find out the value of R

Total resistance of the circuit = (1/2 +1/R) + 4

Emf = 2V

r = 3 ohm

I = 0.25A

Emf = I (R total + r)

2 = 0.25 (Rtotal + 3)

R total = 5 ohm

Therefore,

(1/2 + 1/R) + 4 = 5

1/R = 1- ½

R = 2 ohm

Now for potential difference across 2 ohm resistor

Resultant resistance of 2 ohm resistor as two 2 ohm resistors are connected in parallel

= 1 ohm

V = IR

V = 0.25 x 1

V = 0.25 V

**Question: 29**

**Three resistors of 6.0 ohm, 2.0 ohm and 4.0 ohm are joined to an ammeter A and a cell of emf 6.0 V as shown in figure. Calculate:**

**(a.) the efective resistance of the circuit.**

**(b.) the reading of ammeter**

Solution: Given

R1 = 6 ohm

R2 = 2 ohm

R3 = 4 ohm

a.) As R2 and R3 are connected in series

R’ = 2+4 = 6 ohm

And R’ and R1 are connected in parallel

Total resistance of the circuit, 1/R = 1/6 +1/6

1/R = 2/6

R = 3 ohm

b.) Potential difference, V = 6 V

V = IR

I = V / R

I = 6 /3

I = 2 A

**Question: 30**

**The diagram below in Fig., shows the arrangement of five different resistances connected to a battery of e.m.f. 1.8 V. Calculate:**

**a.) The total resistance of the circuit**

**b.) The reading of ammeter A.**

**Solution:**

The resistance of the resistor that are parallel in the upper part of the circuit

1/R’ = 1/10 + 1/40

1/R’ = 5 /40

R’ = 8 ohm

The resistance of the resistor that are parallel in the lower part of the circuit

1/R’’ = 1/20 + 1/30 + 1/60

1/R’’ = 6 /60

R’ = 10 ohm

Total resistance of the circuit = 10 + 8 = 18 ohm

For reading of ammeter

Current, I = V/R

I = 1.8 /18

I 0.1 A

**Question: 31**

**A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in fig.**

**Find:**

**(a) The reading of the ammeter,**

**(b) The potential difference across the terminals of the cells, and**

**(c) The potential difference across the 4.5 ohm resistor.**

**Solution:** Given

Emf = 2 V

Internal resistance, r = 1.2 ohm

Resistance of the ammeter = 0.8 ohm

The resultant resistance of the resistors connected in parallel

1/R1 = 1/4.5 + 1/9 =3/9

R1 = 3 ohm

Total resistance of the circuit, R= 3 + 0.8 + 1.2 = 5 ohm

**a.) For reading of ammeter **

Current, I = V / R

I = 2 / 5

I = 0.4 A

**b.) Potential difference across the cell **

V1 = emf – Ir

V1 =2 – 0.4 x 1.2

V1 = 1.52 V

**c.) The potential difference across 4.5 ohm resistor **

V = V1 – IR

V = 1.52 – 0.4 x 0.8

V = 1.2 V

**Exercise 8 (C)**

**Question: 1**

**Write an expression for the electrical energy spent in flow of current through an electrical appliance in terms of current, resistance and time.**

**Solution:**

Electrical energy

W = I^{2}R

Where,

W = electrical power

I = current and

R = resistance

**Question: 2**

**Write an expression for the electrical power spent in flow of current through a conductor in terms of (a) resistance and potential difference, (b) current and resistance.**

**Solution:**

a.) Electrical power, P = V^{2}/ R

Where, V = potential difference and R = resistance.

b.) Electrical power, P = I^{2}R

Where, I = current and R = resistance.

**Question: 3**

**Electrical power P is given by the expression P = (Q × V) ÷ time**

**(a.) What do the symbols Q and V represent?**

**(b.) Express the power P in terms of current and resistance explaining the meanings of symbols used their in.**

**Solution:**

a.) Q represents charge and V represents potential difference.

b.) Electrical power, P = I^{2}R

Where, I = current and R = resistance.

**Question: 4**

**Name the S.I. unit of electrical energy. How is it related to Wh?**

**Solution:** S.I. unit of electrical energy is joule.

1 W = 3600 J

**Question: 5**

**Explain the meaning of the statement ‘the power of an appliance is 100 W’.**

**Solution:** power = work done / time

Hence, 100 W energy was consumed in 1 second.

**Question: 6**

**State the S.I. unit of electrical power.**

**Solution:** S.I. unit of electrical power is watt.

**Question: 7**

**(i.) State and define the household unit of electricity.**

**(ii.) What is the voltage of the electricity that is generally supplied to a house?**

**(iii.) What is consumed while using different electrical appliances, for which electricity bills are paid? **

**Solution:**

i.) Unit of household electricity is kilowatt/hr. which means 1 kilowatt energy consumed by household appliances in an hour is called as 1 kilowatt/hr energy.

ii.) 220 -240 volts is generally supplied to a house.

iii.) Electrical energy.

**Question: 8**

**Name the physical quantity which is measured in (i) kW, (ii) kWh. (iii) Wh**

**Solution:**

i.) Electrical power is measured in kW.

ii.) Electrical energy is measured in kW/h

iii.) Electrical energy is measured in Wh.

**Question: 9**

**Define the term kilowatt – hour and state its value in S.I. unit.**

**Solution:** 1 kilowatt/ hour is 1 kilowatt energy consumed by appliances within an hour.

Vale of kilowatt/hour in S.I. unit = 3.6 x 10^{6} J.

**Question: 10**

**How do kilowatt and kilowatt-hour differ?**

**Solution:** Kilowatt hour is the unit of electrical energy and kilowatt/ hour is the unit of electrical power.

**Question: 11**

**Complete the following:**

**(a.) 1 kWh = (1 volt × 1 ampere × ……..) / 1000**

**(b.)** **1 kWh= ________ J**

**Solution:**

a.) 1 hour

b.) 3.6 x 10^{6} J.

** **

**Question: 12**

**What do you mean by power rating of an electrical appliance? How do you use it to calculate (a.) the resistance of the appliance and (b.) the safe limit of the current in it, while in use?**

**Solution: The power rating of an electrical appliance is given buy its power consumption and voltage. **

**a.) Resistance of the appliance **

R= V^{2}/P

Voltage = V

Power = P

**b.) Safe limit of current while in use **

I = P /V

P = power

V = voltage.

**Question: 13**

**An electric bulb is rated ‘100 W, 250 V’. What information does this convey?**

**Solution:**

The power of the bulb is 100 W and it lights up when supplied with voltage of 250 V.

**Question: 14**

**List the names of three electrical gadgets used in your house. Write their power, voltage rating and approximate time for which each one is used in a day. Hence find the electrical energy consumed by each in a month of 30 days.**

**Solution:**

Appliance | Power
In watt |
Voltage, V | Time, hours | Electrical energy
E = P x t |

Led bulb | 18 | 220 | 12 | 0.22 kWh |

Television | 120 | 220 | 5 | 0.6 kWh |

Geyser | 1000 | 220 | 1 | 1 kWh |

**Question: 15**

**Two lamps, one rated 220 V, 50 W and the other rated 220 V, 100 W, are connected in series with mains of voltage 220 V. Explain why does the 50 W lamp consume more power.**

**Solution:** resistance of 50 W bulb

Resistance = V^{2}/ P

R1 = 220 x 220 / 50

R1 = 968 ohm

Resistance of 100 W bulb

Resistance = V^{2}/ P

R2 = 220 x 220 / 100

R2 = 484 ohm

As it stated above the resistance of the 50 W bulb is less than100 W bulb hence it consumed more power, as power is directly proportional to resistance.

**Question: 16**

**Name the factors on which the heat produced in a wire depends when current is passed in it, and state how does it depend on the factors stated by you.**

**Solution:**

a.) amount of heat produced in wire is directly proportional to the square of the amount of current flowing through it.

b.) The amount of heat produced by a wire is directly proportional to the resistance of the wire

c.) And lastly amount of heat is directly proportional to the time for which current flows through the wire.

**MULTIPLE CHOICE TYPE**

**Question: 1**

**When a current I flows through a resistance R for time t, the electrical energy spent is:**

(a.) IRt

(b.) I^{2}Rt

(c.) IR^{2}t

(d.) I^{2}R / t

**Solution:** (b) I^{2}Rt

**Question: 2**

**An electrical appliance has a rating 100 W, 120 V. The resistance of element of appliance when in use is:**

(a) 1.2 ohm

(b) 144 ohm

(c) 120 ohm

(d) 100 ohm

**Solution:** (b) 144 ohm

**NUMERICALS**

**Question: 1**

**An electric bulb of resistance 500 ohm draws current 0.4 A from the source. Calculate: (a) the power of bulb and (b) the potential difference at its end.**

**Solution:** Given

Resistance, R= 500 ohm

Current, I = 0.4 A

Potential difference, V = I x R

V = 0.4 x 500

V = 200 V

Power, P = V x I

P = 200 x 0.4

P = 80 W

**Question: 2**

**A current of 2 A is passed through a coil of resistance 75 Ω for 2 minutes. (a) How much heat energy is produced? (b) How much charge is passed through the resistance?**

**Solution:** Given

Resistance, R= 75 ohm

Current, I = 2 A

Time, t = 2 min = 120 s

Heat produced = I^{2}Rt

= 2 x 2 x 75 x120

= 36000 J

Charge passed = I x t

Q = 2 x 120

Q = 240 C.

**Question: 3**

**Calculate the current through a 60 W lamp rated for 250 V. If the line voltage falls to 200 V, how is power consumed by the lamp affected?**

**Solution:** Given

Power, P = 60 W

Voltage, V = 250 V

As we know,

R = V^{2}/ P

When the voltage drops 200 V

Power = V^{2}/ R

= 200 x 200 / 250 x 250 / 60

Power reduced = 38.4 W

**Question: 4**

**An electric bulb is rated ‘100 W, 250 V’. How much current will the bulb draw if connected to a 250 V supply?**

Solution: Given

Power, P = 100 W

Voltage, V = 250 V

Current, I = P / V

I = 100 / 250

I = 0.4 A

**Question: 5**

**An electric bulb is rated at 220 V, 100 W. (a) What is its resistance? (b) What safe current can be passed through it?**

**Solution:** Given

Power, P = 100 W

Voltage, V = 220 V

Resistance, R = V^{2} / P

R = 220 x 220 / 100

R = 484 ohm

Safe limit of current

I = P / V

I = 100 / 220

I = 0.45 A

**Question: 6**

**A bulb of power 40 W is used for 12.5 h each day for 30 days. Calculate the electrical energy consumed.**

**Solution:** Given

Power, P = 40 W

Time, t= 12.5 hours

Energy consumed = P x t

E = 40 x 12.5

E = 500 Wh

Energy consumed in 30 days

E = 500 x 30

E = 15000 Wh = 15 kWh

**Question: 7**

**An electric press is rated ‘750 W, 230 V’. Calculate the electrical energy consumed by the press in 16 hours**

**Solution:** Given

Power, P = 750 W

Time, t = 16 hours

Energy consumed = 750 x 16

E = 12000 Wh = 12 kWh

**Question: 8**

**An electrical appliance having a resistance of 200 ohm is operated at 200 V. Calculate the energy consumed by the appliance in 5 minutes (i) in joule, (ii) in kWh**

**Solution:** Given

Resistance, R = 200 ohm

Voltage, V = 200 V

Time, t = 5 min = 300 s

Energy = V^{2}t /R

E = 200 x 200 x 300 / 200

E = 60000 J

As we know,

3.6 x 10^{6}J = 1 kWh

60000 J = 60000 /3.6 x 10^{6}

= 0.0167 kWh

**Question: 9**

**A bulb rated 12 V, 24 W operates on a 12 volt battery for 20 minutes. Calculate:**

**(i) the current flowing through it, and**

**(ii) the energy consumed.**

**Solution:** Given

Power, P = 24 W

Voltage, V = 12 V

Time, t = 20 min = 120 s

Current, I = P / V

I = 24/12

I = 2 A

Energy = P x t

= 24 x 120

= 28,800 J

**Question: 10**

**A current of 0.2 A flows through a wire whose ends are at a potential difference of 15 V. Calculate:**

**(i.)The resistance of the wire, and**

**(ii.)The heat energy produced in 1 minute.**

**Solution:** Given

Current, I = 0.2 A

Potential difference, V = 15 V

Resistance, R = V / I

R = 15 / 0.2

R = 75 ohm

Heat produced in 1 min

Heat produced = I^{2}Rt

= 0.2 x 0.2 x 75 x 60

= 180 J

**Question: 11**

**What is the resistance, under normal working conditions, of an electric lamp rated at ‘240 v’, 60 W? If two such lamps are connected in series across a 240 V mains supply, explain why each one appears less bright.**

**Solution:** Given

Voltage, V = 240 V

Power, P = 60 W

Resistance, R = V^{2} / P

R = 240 x 240 / 60

R = 960 ohm

Current, I = V / R

I = 240 / 960

I = 0.25 A

Now the bulbs are connected in series then resultant resistance becomes

960 + 960 = 1920 ohm

Therefore, current = 240 / 1920

Current = 0.125 A

As the current flowing through the bulbs is reduced they appear less bright.

**Question: 12**

**Two bulbs are rated 60 W, 220 V and 60 W, 110 V, respectively. Calculate the ratio of their resistances.**

**Solution:** Given

Power, P1 = 60 W

Voltage, V1 = 220V

Resistance, R1 = V^{2} / P

= 220 x 220 / 60

Power, P2 = 60 W

Voltage, V2 = 110 V

Resistance, R2 = V^{2} / P

R2 = 110 x 110 / 60

Therefore,

R1/R2 = 220 x 220 /60 / 110 x 110 /60

R1/R2 = 220 x 220 / 110 x 110

R1/R2 = 4/1

**Question: 13**

**An electric bulb is rated 250 W, 230 V.**

**(i.) the energy consumed in one hour, and**

**(ii.) the time in which the bulb will consume 1.0 kWh energy when connected to 230 V mains?**

**Solution:** Given

Power, P = 250 W

Voltage, V = 230 V

Time, t = 1 hours = 3600s

Energy consumed =250 x 3600

E = 9 x 10^{5} J

Energy = 1 kWh

1000 Wh = power x time

Time = 1000 / 250

Time = 4 hours.

** **

**Question: 14**

**Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply. Calculate:**

**(i.) The total current taken from the supply,**

**(ii.) The resistance of each heater, and**

**(iii.) The energy supplied in kWh to the three heaters in 5 hours.**

**Solution:** Given

Voltage, V = 100 V

Power, P = 250 W

Current, I = P / V

I = 250 / 100

I = 2.5 A

Therefore, current taken by three heaters = 3 x 2.5 = 7.5 A

Resistance of each heater,

Resistance, R = V^{2} / P

R = 100 x 100 / 250

R = 400 ohm

Energy consumed = P x t

E = 250 x 5

E = 1250 Wh

E = 1.25 kWh

Therefore, energy of three heaters = 3 x 1.25

= 3.75 kWh.

**Question: 15**

**A bulb is connected to a battery of p.d. 4 V and internal resistance 2.5 ohm. A steady current of 0.5 A flows through the circuit. Calculate:**

**(i.) The total energy supplied by the battery in 10 minutes,**

**(ii.) The resistance of the bulb, and**

**(iii.) The energy dissipated in the bulb in 10 minutes.**

**Solution:** Given

Resistance of the battery, r = 2.5 ohm

Voltage, V = 4 V

Current, I = 0.5 A

Time, t = 10 min = 600 s

Resistance, R = V / I

R = 4 / 0.5

R = 8 ohm

Energy = V^{2}t /R

E = 4 x 4 x 600 / 8

E = 1200 J

As total resistance is 8 ohm

Resistance of the bulb is 5.5 ohm.

Energy of bulb (heating) = I^{2}Rt

= 0.5 x 0.5 x 5.5 x 600

= 825 J.

**Question: 16**

**Two resistors A and B of 4 ohm and 6 ohm, respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate: (i) the power supplied by the battery, (ii) the power dissipated in each resistor.**

**Solution:** Given

Resistance connected in parallel resultant resistance

1/ R = ¼ + 1/6 = 10 /24

R = 2.4 ohm

Voltage, V = 6 V

Power supplied by the battery = V^{2}/ R

P = 6 x 6 / 2.4

P = 15 W

Power at resistor of 4 ohm = V^{2}/ R

= 36 / 4

= 9 W

Power at resistor of 6 ohm = V^{2}/ R

= 36 / 6

= 6 W

**Question: 17**

**A battery of e.m.f. 15 V and internal resistance 2 ohm is connected to two resistors of resistances 4 ohm and 6 ohm joined in series. Find the electrical energy spent per minute in 6 ohm resistor.**

**Solution:** Given

Voltage, V = 15 V

Internal resistance, r = 2 ohm

Time, t = 1 min = 60s

Resultant resistance of two resistance in series and battery = 4 + 6 + 2 = 12 ohm.

Current, I = V / R

I = 15 / 12

I = 1.25 A

Voltage across 6 ohm resistor

V = I x R = 1.25 x 6 = 7.5 V

Energy = V^{2}t / R

E = 7.5 x 7.5 x 60 / 6

E = 562.5 J

**Question: 18**

**Water in an electric kettle connected to a 220 V supply took 5 minutes to reach its boiling point. How long will it take if the supply had been of 200 V?**

**Solution:** Given

Voltage, V1 = 220 V

Voltage, V2 = 200 V

Time, t1 = 5 min

Heat energy = (V^{2 }/ R)t

As heat gained to boil water is same

(V1^{2}/R)t1= (V2^{2}/ R)t2

t2 = (220 x 220 /200 x 200) x 5

t2 = 6.05 min.

**Question: 19**

**An electric toaster draws current 8 A in a circuit with source of voltage 220 V. It is used for 2 h. Find the cost of operating the toaster if the cost of electrical energy is ₹ 4.50 per kWh**

**Solution:** Given

Current, I = 8 A

Voltage, V = 220 V

Time, t = 2 h

Energy used = P x t

E = 220 x 8 x 2

E = 3520 Wh = 3.52 kWh

Cost of operating = 3.52 x 4.50

= 15.84 rupees

**Question: 21**

**A geyser is rated 1500 W, 250 V. This geyser is connected to 250 V mains. Calculate:**

(i)The current drawn,

(ii)The energy consumed in 50 hours, and

(iii)The cost of energy consumed at ₹ 4.20 per kWh.

**Solution:** Given

Power, P = 1500 W

Voltage = 250 V

Current, I = P / V

I = 1500 / 250

I = 6 A

Energy consumed in 50 hours

Energy = P x t

E = 1.5 x 50 = 75 kWh

Cost of use = 75 x 4.20

= 315 rupees.