NCERT Solutions Class 10 Science Chapter 12 Electricity
NCERT Solutions Class 10 Science Chapter 12 Electricity: National Council of Educational Research and Training Class 10 Science Chapter 12 Solutions – Electricity. NCERT Solutions Class 10 Science Chapter 12 PDF Download.
NCERT Solutions Class 10 Science Chapter 12: Overview
Board |
NCERT |
Class |
10 |
Subject |
Science |
Chapter |
12 |
Chapter Name |
Electricity |
Topic |
Exercise Solutions |
NCERT Solutions Class 10 Science Chapter 12 – Electricity
Middle Exercise:
Page No. 200
1.) What does an electric circuit mean?
Ans:
Electric circuit is the continuous and closed path of an electric current. If the circuit is broken at anywhere then there will be current flowing is stopped and bulb doesn’t glow.
2.) Define the unit of current?
Ans:
- Current is the rate of flow of charge.
- The SI unit of current is A.
- One ampere is the current due to the flow of one coulomb charge per second.
Thus,
1A = 1C/ 1s
3.) Calculate the number of electrons constituting one coulomb of charge?
Ans:
- 1C = 1.6*10-19 * charge on one electron
- Number of electrons = 1/1.6*10-19 = 0.625*1019 electrons
- Thus, number of electrons = 6.25*1018
- Thus, 6.25*1018 electrons constitute the 1 coulomb of charge.
Page 202
1.) Name the device that helps to maintain a potential difference across a conductor?
Ans:
A battery which is made up of group of cells is the device which helps to maintain a potential difference across a conductor. Also, a single cell is also used to maintain potential difference across a conductor.
2.) What is meant by saying that the potential difference between two points is 1V?
Ans:
- The SI unit of potential difference is volt.
- And the potential difference between two points is 1V means that 1J of work is done in moving the charge of one coulomb from one point to another point.
- Hence, 1V = 1J/ 1C
3.) How much energy is given to each coulomb of charge passing through a 6V battery?
Ans:
Given that,
V= 6 V
We know that,
W = VQ = 6*1 = 6J
Thus, 6J of energy will be given to each coulomb of charge passing through a 6V battery.
Page 209
1.) On what factors the resistance of the conductor depends?
Ans:
The resistance of the conductor depends on
- length of the conductor
- area of cross section of the conductor
- and the nature of the material used.
Thus,
- Resistance of he conductor is given by,
R= qL/A
2.) Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source?
Ans:
- As we know that, resistance of the conductor is inversely proportional to the area of cross section of the conductor.
- More the area less will be the resistance and more current will flows through the conductor.
- As the wire is thinner it’s cross section of area increases and hence resistance decreases die to which current flows through thin wire easily.
3.) Let the resistance of an electrical component remains constant while the potential difference across the two ends of the components decreases to half of its former value. What change will occur in the current through it?
Ans:
- Given that, resistance of an electrical component remains constant but it’s voltage become half of its former value.
- Hence, by Ohm’s law
V= IR
Then, I = V/ R
- Now, the new voltage is V’ = V/2
- Thus, current through it after change in voltage will be,
I = (V/2)/ R = 1/2 *(V/R) = I/2
- Thus, it is seen that current flowing also half of its former value.
4.) Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Ans:
- The conductors of heating devices such as bread-toasters and electric irons are generating more heat and heat generated is directly proportional to the resistivity of the material used.
- And we know that, alloys has more resistivity than the pure metals. Hence, alloys are used rather than pure metal in conductors of electric heating devices.
5.) Use the data in table 12.2 to answer the following:
a) Which among iron and mercury is better conductor?
b) Which material is the best conductor?
Ans:
As we know that more is the resistivity the materials are bad conductors of electricity and less is the resistivity the materials would be good conductors of electricity.
a)
From table, the resistivity of iron is less as compared to the mercury. Hence iron is better conductor than mercury.
b)
- The material having lowest resistivity is the best conductor.
- From table, silver is the material having lowest resistivity which is 1.60*10-8 ohm m.
Hence, silver is the best conductor of electricity.
Page 213
1.) Draw a schematic diagram of a circuit consisting of a battery of three cells of 2V each, a 5ohm resistor, 8 ohm resistor and a 12 ohm resistor and a plug key, all connected in series.
Ans:
2.) Redraw the above-mentioned circuit putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 ohm resistor. What would be the readings in the ammeter and voltmeter?
Ans:
- The voltmeter is connected in parallel to the resistor of resistance 12 ohm then the reading shown by voltmeter will be the voltage drop across the 12 ohm resistor only.
- And we connected the ammeter in series with the circuit as shown in figure. Thus, it gives the total current flowing through all the resistors.
- Total resistance of the circuit = 5+8+12= 25 ohm
- By, Ohm’s law
- V= IR
- I= V/ R = 6/25= 0.24 A
- Thus, 0.24 A is the current flowing through each resistor.
- Hence, the voltage drop across the 12 ohm resistor will be,
- V’= IR = 0.24*12= 2.88 V
- Thus, in this case the reading shown by ammeter will be 0.24 A and the reading shown by voltmeter will be 2.88 V.
Page 216
1.) Judge the equivalent resistance when the following are connected in parallel
a) 1 ohm and 106 ohm
b) 1 ohm, 103 ohm and 106 ohm.
Ans:
a)
- When the resistances 1 ohm and 106 ohm are connected in parallel then the equivalent resistance will be given by,
1/Re = 1/1 + 1/106 =( 1+ 10-6)~ 1
Thus, 1/Re = 1
Hence, Re= 1 ohm
- When the resistances 1 ohm and 106 ohm are connected in parallel then equivalent resistance will be 1 ohm.
b)
- When the resistances 1 ohm, 103 ohm and 106 ohm are connected in parallel then the equivalent resistance will be given by,
1/Re = 1/1 + 1/103 + 1/106
106/Re = 106 + 103 + 1,001,000
Thus, 106/Re = 1,001,00
- Hence, Re/ 106 = 1/1,001,000
Re= 0.999 ohm
- Thus, when the resistances 1 ohm, 103 ohm and 106 ohm are connected in parallel then the equivalent resistance will be 0.999 ohm.
2.) An electric lamp of 100 ohm, a toaster of resistance 50 ohm and a water filter of resistance 500 ohm are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Ans:
- Given that, an electric lamp of resistance 100 ohm, a toaster of resistance 50 ohm and a water filter of resistance 500 ohm are connected in parallel across the 220 V supply.
- Hence, the equivalent resistance of all the resistors connected in parallel will be,
1/Re = 1/100 + 1/50 + 1/500 =
500/Re = 5 + 10 + 1 = 16
500/ Re = 16
- Re = 500/16 ohm
- So the maximum current flowing will be,
V= I R
I = V/ R = 220/ (500/16) = 7.04 A
- This is the maximum current which is drawn by the electric iron.
- Hence, by Ohm’s law the resistance of the electric iron will be,
R’ = V/ I = 220/ 7.04 = 31.25 ohm
- Thus, the maximum current drawn by the electric iron will be 7.04 A and the resistance of the electric iron will be 32.25 ohm
3.) What are the advantages of connecting electric devices in parallel with the battery instead of connecting them in series?
Ans:
- The total effective resistance of resistors in parallel combination becomes less or less than each individual resistance connected in parallel.
- Due to which the current through each component in parallel combination is also different.
- In parallel combination, if any one of the component get damaged then the whole circuit not get damaged. We can make in parallel combination one component ON and other OFF at a time.
- 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 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 through each branch will be different due to parallel combination.
- Also, if we want to turn OFF any one device and other have to make ON then we have to use parallel combination only. And if any one of the device get damaged due to extra current then that device stopped only and whole circuit will be working and this is possible only in parallel com
- Because of the above all advantages, the parallel combination is used in domestic wiring.
4.) How can three resistors of resistances 2ohm, 3 ohm and 6 ohm be connected to give a total resistance of
a) 4 ohm and b) 1 ohm
Ans:
Given three resistors are having resistances 2 ohm, 3 ohm and 6 ohm.
a)
- Now we connect the resistors of resistances 3 ohm and 6 ohm in parallel and to this combination we connect a resistor of resistance 2 ohm in series and we find the equivalent resistance.
1/Re = 1/3 + 1/6 = 3/6 = 1/2
Hence, Re = 2
- Total resistance will be R = 2 + Re = 2 + 2 = 4 ohm
b)
- Now we connect all the resistors in parallel combination and we find the equivalent resistance of the circuit.
1/Re = 1/2 + 1/3 + 1/6 = (3+2+1)/6 = 6/6 = 1
- Hence, Re = 1ohm
5.) What is a) the highest b) the lowest total resistance that can be secured by combinations of four coils of resistance 4ohm, 8ohm, 12 ohm, 24 ohm?
Ans:
- We know that, when the resistors are connected in series then there will be highest resistance.
- Hence, if the resistances 4 ohm, 8 ohm, 12 ohm and 24 ohm are connected in series then the equivalent resistance of them is the highest resistance which is given by,
Re = 4 + 8 + 12 + 24 = 48 ohm
This is the highest resistance.
- And when the resistances are connected in parallel then their equivalent resistance will be the lowest resistance.
- If the resistances 4 ohm, 8 ohm, 12 ohm and 24 ohm are connected in parallel then equivalent resistance will be given by,
1/Re = 1/4 + 1/8 + 1/12 + 1/24
24/Re = 6 + 3 + 2 + 1 = 12
Thus, 24/Re = 12
Hence, Re/24= 1/12
Re= 24/12 = 2 ohm.
This is the lowest resistance.
Page 218
1.) Why does the cord of an electric heater not glow while the heating element does?
Ans:
- As we know that, the heating element in an electric heater is made up of alloys which are having high resistance.
- As the current is passed through these alloys then large amount of heat is generated due to which the heating element become hotter and hence it glows red.
- While the cord of the electric heater is made from copper which is having low resistance and hence it doesn’t glow.
2.) Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference if 50 V.
Ans:
Given that,
Charge Q= 96000 coulomb
Time t = 1 hr = 60*60= 3600 second
Potential difference V = 50V
Heat generated is given by,
H= VIt
- And we know that,
I = Q/t = 96000/ 3600 = 26.66 = 27 A
- Now, the heat generated is given by,
H = VIt = 50*27*3600= 4,860,000 J
= 4.86*106 J
Thus the heat generated will be 4.86*106 J
3.) An electric iron of resistance 20 ohm takes a current of 5A. Calculate heat developed in 30s.
Ans:
Given that,
R= 20 ohm
I= 5A
Time t= 30s
We know that, the heat produced will be given by,
H= I2Rt = 25*20*30= 500*30= 15000 = 1.5*104 J
Thus, the heat generated in 30s by the electric iron will be 1.5*104J.
Page 220
1.) What determines the rate at which energy is delivered by a current?
Ans:
The rate at which energy is delivered by a current determines the power of that electric device.
2.) An electric motor takes 5A from a 220V line. Determine the power of the motor and the energy consumed in 2 hr?
Ans:
Given that,
Current I = 5A
Voltage V = 220V
Time t= 2hr = 2*3600= 7200 second
The power is given by,
P= VI= 220*5=1100 W
And the heat generated is given by,
H= P*t = 1100*7200= 7920000 J
= 7.92*106 J
Thus, the power will be 1100 W and energy consumed will be 7.92*106 J
Exercise
1.) A piece of water of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’ then the ratio R/R’ is
- a) 1/25
- b) 1/5
- c) 5
- d) 25
Ans: d) 25
Because,
- As we know that,
- The resistance of conductor is directly proportional to the length of the conductor.
- Hence, if the wire having resistance R is cut into five parts then the resistance of each part will be R/5.
- These parts are then connected in parallel, hence their equivalent resistance R’ will be,
1/R’ = 5(5/R) = 25/R
Thus, R/R’= 25
2.) Which of the following term does not represent electrical power in a circuit?
a) I2R
b) IR2
c) VI
d) V2/R
Ans: b) IR2
- Because, we know that power of an electrical circuit is given by,
P= VI
But, V= IR
- Then, P = (IR)I = I2R
- Now, I = V/R
- So, P= V(V/R) = V2/R
- So, options a), c) and d) are correct and only option b) is not correct.
3.) An electric bulb is rated 220V and 100W. When it is operated on 110V, the power consumed will be
a) 100W
b) 75W
c) 50W
d) 25W
Ans: d) 25W
- We know that, power of an electrical circuit is given by,
P= V2/R
- Hence, R= V2/P
- Given that, P= 100W, V= 220V
- Thus, R = 220*220/100= 484 ohm
- Now, it is operated on 110V the power consumed will be,
P = V2/R = 110*110/484
P= 25W
4.) Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combination would be
a) 1:2
b) 2:1
c) 1:4
d) 4:1
Ans: c) 1:4
- Let R be the resistance of each wire.
- When the wires are connected in series then the equivalent resistance will be R1 = R + R = 2R
- And when wires are connected in parallel then equivalent resistance will be
1/R2 = 1/R + 1/R = 2/R
Thus, R2 = R/2
- The heat produced will be given by,
H = V2t/R
- The ration of heat produced in series combination Hs and heat produced in parallel combination Hp is given by,
- Hs/Hp = (V2t/R1)/ (V2t/R2)
= R2/R1 = (R/2)/2R = 1/4
- Thus, the ratio of heat produced in series and parallel combination will be 1:4.
5.) How is a voltmeter connected in the circuit to measure the potential difference between two points?
Ans:
Voltmeter is used to measure the potential difference between any two points in the circuit. To measure potential difference the voltmeter should be connected in parallel to the two points in the circuit.
Alternative Answer –
Connect the voltmeter parallel to the circuit to measure the potential difference between two points.
6.) A copper wire has diameter 0.5 mm and resistivity of 1.6*10-8 ohm m. What will be the length of this wire to make its resistance 10 ohm? How much does the resistance change if the diameter is doubled?
Ans:
Given that,
Diameter of copper wire = 0.5 mm
Hence, it’s radius will be r = 0.5/2 = 0.25 mm = 0.025 cm
And resistance of wire R = 10 ohm.
Resistivity of the wire = 1.6*10-8 ohm m = 1.6*10-6 ohm cm.
- The resistance of the wire in terms of resistivity of the wire is given by,
R = qL/A
- But wire is in the cylindrical form whose area will be πr2.
- So R = qL/ πr2
- Thus length of wire having resistance 10 ohm is given by,
L= πr2*R/q = (3.14*0.025*0.025*10)/1.6*10-6
= 0.019625/ 1.6*10-6
= 0.01226*106
= 122*102cm
= 122 m
- Thus, when the resistance of the wire is 10 ohm then the length of the wire should be 122m.
We know that,
- Resistance of the wire is given by,
R = qL/A
= qL/ πr2
= 4qL/ πd2
That means, R α 1/d2
- Thus, when the diameter of the wire is doubled then resistance of the wire should be,
R α 1/4d2
- Thus, the new resistance will be the one fourth of original resistance of the wire.
Thus,
New resistance= 10/4 = 2.5
- And change in resistance will be given by,
10 – 2.5 = 7.5 ohm
- Thus, when the diameter of the wire is doubled the change in resistance will be 7.5 ohm.
Alternative Answer –
Given that,
A copper wire of diameter, d = 0.5 mm = 5×10-4 m
Resistivity of the copper wire, ρ = 1.6×10-8 Ω m
Resistance of the copper wire, R1 = 10 Ω
w.k.t Resistance, R1 = ρ ( l / A)
Where l is the length of the copper wire and A is the Area of cross section of the copper wire.
A = π ( d/ 2)2
l = R1 A /ρ
l = ( R1 × π ( d/ 2)2. ) / ρ
Substitute the above values in the equation
l = (10 × 3.142 ( ( 5 × 10-4 )/ 2)2)/ 1.6× 10-8
= ( 31.42 × ( 2.5 × 10-4 )2 ) / 1.6 × 10-8
= 19.63 × 108 × 6.25 × 108
l = 122.68 m
Therefore the length of the wire is 122.68 m
Let us consider, changed resistance is is R2 = ρ ( l / A)
Now we find changed resistance if the diameter is double, d × 2 = 5 × 10-4 × 2
d2 = 10 × 10-4 = 10-3 m
R2 = ((1.6×10-8)(122.68))/(( ½ × 10-3)2 × 3.142)
= (196.288×10-8)/(0.25×10-6×3.142)
= (196.288×10-8×106)/(0.7855)
= 249.88×10-2
R2 =2.4988Ω
Therefore the length of the copper wire is 122.6 m and the changed resistance is 2.4988Ω
7.) The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:
Plot a graph between V and I and calculate resistance of that resistor.
Ans:
- The V -I graph is as shown below which is a straight line passing through origin. Which means current and voltage are varying linearly with each other.
- And the slope of the graph is constant which is given by,
- Slope = change in voltage/ change in current = resistance
- Thus, we have taken two points A and B from which we can determine the resistance of the resistor used.
- Hence,
Slope = (10.2 – 5)/ (3 – 1.5) = 5.2/ 1.5 = 6.86 ohm
- Thus, the resistance of the resistor will be 6.86 ohm.
Alternative Answer –
From the graph slope = BC/AC
= 2/6.8
= 0.2941
w.k.t Resistance,R = 1/slope
R = 1/ 0.2941
R = 3.431Ω
Therefore the resistance of the resistor is 3.431Ω
8.) When a 12V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Ans:
Given that,
Voltage V= 12 V
Current I = 2.5 mA = 2.5*10-3 A
- According to Ohm’s law,
V = I*R
- Then, R= V/I = 12/ 2.5*10-3 = 4.8*103 ohm
- Thus, the resistance of the resistor will be 4.8*103
Alternative Answer –
Given that,
The voltage of the battery,V = 12 V
The current flow through the circuit, I = 2.5 mA = 2.5 × 10-3 A
Using Ohm’s law,
V = IR
R = V/ I
= 12 / 2.5 × 10-3
R = 4.8 × 103 Ω
Therefore the resistance of the resistor is 4.8 KΩ.
In case you are missed :- Previous Chapter Solution
9.) A battery of 9 V is connected in series with resistors of 0.2 ohm, 0.3 ohm, 0.4 ohm, 0.5 ohm and 12 ohm respectively. How much current would flow through the resistor 12 ohm?
Ans:
- Given that, resistors having resistance 0.2, 0.3, 0.4, 0.5 and 12 ohm are connected in series.
- Then equivalent resistance of the circuit is given by,
Re = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 ohm
- Given that, the potential difference applied is 9V.
Thus, V= IR
I = V/ R = 9/ 13.4 = 0.67 A
- But, we know that when the resistors are connected in series combination then the current through each resistor connected in series will be same.
- Hence, here the current flowing through resistor 12 ohm will be 0.67 A.
Alternative Answer –
Given that,
The voltage of the battery, V = 9 V
and resistor which are connected in series combination,
R1= 0.2Ω. R2= 0.3Ω. R3= 0.4Ω. R4= 0.5Ω. R5= 12Ω
w.k.t when resistors we connect in series combination then equivalent resistance is, Req = R1 + R2 +R3 + R4 +R5
= 0.2 +0.3 + 0.4+ 0.5+ 12
Req = 13.4Ω
Now using Ohm’s law,
V = IReq
I = V/ Req
= 9/ 13.4
I = 0.671 A
Therefore when 9V battery connected with series combination of resistance then 0.671 A current is flow through the 12Ω resistor.
10.) How many 176 ohm resistors in parallel are required to carry 5A on a 220V line?
Ans:
- Given that, 5A is the current flowing when potential difference applied is 220V.
- Then we can find the total resistance of the circuit by using Ohm’s law,
V = IR
R = V/ I = 220/5= 44 ohm
- Thus, the total resistance of the circuit is 44 ohm.
- Let us consider x be the number of resistors each having resistance 176 ohm are connected in parallel then the equivalent resistance obtained will be 44ohm.
- Hence, we can write,
1/44 = 1/176 + 1/ 176 + … up to x times
Hence, 1/44 = x/ 176
So, x= 176/44 = 4
Thus, when we connect 4 resistors each of resistance 176 ohm in parallel across which voltage 220V is applied then current through each resistor will be 5 A.
Alternative Answer –
Given that,
n number of resistance are connected in parallel combination that means
R1 = R2 = R3 =…………= Rn = 176Ω
w.k.t when resistance are connected in parallel combination then,
(1/Req)= (1/R1) + (1/R2) + (1/R3) +……….+ (1/Rn)
(1/Req) = (1/ Rn)
(1/Req) = (n/Rn)
Req = R1/n
Req = 176/n
Using Ohm’s law,
V = IReq
Given that,
Voltage, V = 220 V
Current, I = 5 A
V = I ×( 176/n)
n =( I × 176) V
n =( 5 × 176)/220
n = 880/220
n = 4
There are four 176 ohm resistors (in parallel) are required to carry 5A on a 220 V line.
11.) Show how you would connect three resistors, each of resistance 6 ohm so that the combination has a resistance of a) 9 ohm b) 4 ohm.
Ans:
We have three resistors of resistance 6 ohm.
And we have to connect then in such way that equivalent resistance will be 9 ohm and 4 ohm respectively.
a)
- Now, when we connect two resistors of resistance 6 ohm in parallel and to this parallel combination, we connect a resistance 6ohm in series as shown in figure. Then the equivalent resistance will be,
1/Re= 1/6 + 1/6 = 2/6= 1/3
Re = 3 ohm
So in this case total resistance of the circuit will be R = Re + 6 = 3 + 6 = 9 ohm.
b)
- Now, we connect two resistors of resistance 6 ohm in series’ first and then to this series combination we connect the third resistance in parallel as shown in figure.
Then, Re = 6 + 6 = 12 ohm
Thus, total resistance of the circuit will be,
1/R = 1/12 + 1/6 =3/12 = 1/4
Thus, R = 4 ohm
Alternative Answer –
Given that,
Each resistor has resistance 6Ω
There are 3 resistores, R1 = R2 = R3 = 6Ω
Now first we connect three resistors in parallel combination
Then (1/Req) = (1/R1) + (1/R2) + (1/R3)
(1/Req) = (⅙) + (⅙) + (⅙)
= ((6+6)/(6×6)) + (⅙)
= ⅓ + ⅙
(1/Req) = ½
Req = 2 which is not a required answer.
Next we connect three resistors in series combination then,
Req = R1+ R2 +R3
= 6+ 6+ 6
Req =18Ω which is also not required answer.
Next we try, connect any two resistors in series or parallel.
i) First we connect any two resistors in parallel then,
R1 = 6Ω R2 = 6Ω R3 = 6Ω
Consider R1 and R2 are connected in parallel
(1/Req) = (1/R1) + (1/R2)
= ⅙ + ⅙
= ⅓
Req = 3
Then connect Req and R3 in series then , Resistor R = Req + R3
R = 3 + 6
R = 9Ω
Therefore, to get Resistance 9Ω then we connect any two resistor in parallel and connect Req ( Parallel of R1 and R2 ) and third resistor in series.
ii) Next we connect any two resistors in series,
R1 = 6Ω R2 = 6Ω R3 = 6Ω
Consider R1 and R2 are connected in series, then equivalent resistance is Req = R1 + R2
Req = 6 + 6 = 12Ω
Then connect Req and R3 in parallel then the resistance,
1/R = (1/Req) + (1/R3)
1/R = (1/12) + (⅙)
= ¼
R = 4Ω
Therefore, to get Resistance 4Ω then we connect any two resistor in series and connect Req ( series of R1 and R2 ) and third resistor in parallel.
12.) Several electric bulbs designed to be used on 220V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5A?
Ans:
Given that,
Voltage applied V = 220 V
Power of each bulb P = 10 W
Maximum current flowing I = 5 A.
- We can find the resistance of each bulb from its power.
P = V2/R
Hence, R = V2/ P = 220*220/10 = 4840 ohm.
- Thus, resistance of each bulb is 4840 ohm.
- Now we connect x number of bulbs each having resistance 4840 ohm in parallel then equivalent resistance will be,
1/Re = 1/4840 + 1/4840 + … x times
1/Re = x/4840
Thus, Re = 4840/ x
- But, by using Ohm’s law we can find the total resistance or equivalent resistance of the circuit as,
R = V/ I = 220/5 = 44 ohm
Thus, 4840/x = 44
Hence, x = 4840/44 = 110
- Thus, we have to connect 110 bulbs in parallel to satisfy given condition.
Alternative Answer –
Given that,
Voltage of the electric supply line, V = 220V
Power of each bulb, P= 10 W
Maximum current allowed,I = 5A
First we find resistance of each bulb,
Using the formula, Joules law of heating.
Power, P = V2/ R1
R1 = (220)2/10
= 48400/10
Resistance R1 = 4840Ω
Next we find the equivalent resistance using Ohm’s law,
V = IReq
Req = V/ I
= 220 /5
Req = 44Ω
Next we find the number of lamps.
Given that all lamps are connected in parallel with each other then we use the formula,
1/Req = (1 / R1 )+ (1 / R2 ) +……..+(1 / Rn)
1/Req = n × (1/R1)
n = R1 × Req
= 4840 / 44
n = 110
Therefore 110 lamps are connected to each other in parallel combinations.
13.) A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 ohm resistance, which may be used separately, may in series or in parallel. What are the currents in the three cases?
Ans:
Given that,
Voltage V = 220 V
Resistance of coils A and B = 24 ohm
a)
- When these two coils are used separately then the current through each coil A and B is given by,
I = V/ R = 220/ 24 = 9.167 A
- Thus, when the coils are used separately then current through each coil will be 9.167 A.
b)
- When these two coils are connected in series then the equivalent resistance will be,
Re = 24 + 24 = 48 ohm
- Thus, current through circuit in this case will be,
I = V/ R = 220/ 48 = 4.58A
c)
- When these two coils A and B are connected in parallel then the equivalent resistance will be given by,
1/Re = 1/24 + 1/24 = 2/24 = 1/12
Re = 12 ohm
- Thus, the current through circuit in this case will be,
I = V/R = 220/12 = 18.33 A
Alternative Answer –
Given that
Voltage, V = 220V
and Resistance, R = 24Ω
i) In the first case,when coil used separately
Using Ohm’s law, V = I1R1
I1 = V/R1
I1 = 220/24
I1 = 9.166A
In the case when coil used separately the current is 9.166A
ii) In the second case, when the coil is used in series.
Then Req = R1 +R2
= 24+24
Req = 48Ω
Using Ohm’s law,
V = I2Req
V/Req = I2
I1 = 220/48
I1 = 4.5833A
In the case when the coil used in series then the current is 4.5833A
iii). In the third case, when the coil is used in parallel.
Then 1/Req = (1/R1) + (1/R2)
= (1/24) + (1/24)
= 1/12
Req = 12
Using Ohm’s law,
V = I3R
I3 = V/ R
= 220/12
I3 = 18.33 A
In the case when the coil used in parallel then the current is 18.33 A
14.) Compare the power used in the 2ohm resistor in each of the following case:
a) a 6V battery in series with 1 ohm and 2 ohm resistors and
b) a 4V battery in parallel with 12 ohm and 2 ohm resistor.
Ans:
a)
- When a 6V battery is connected in series with 1 ohm and 2 ohm resistors then, the equivalent resistance of the circuit is given by,
Re = 1 + 2 = 3 ohm
Given that,
V= 6V
- Then according to Ohm’s law the current through the circuit will be,
I= V/Re = 6/3= 2A
- In this case, the power consumed by the resistor of resistance 2 ohm is given by,
P= I2*R = 4*2 = 8 W
- Thus, in this case power consumed by 2-ohm resistor will be 8W.
b)
- In this case, a 4V battery is connected in parallel with 12 ohm and 2 ohm resistors then power consumed by resistor 2 ohm will be,
P = V2/R = 16/8 = 8W
Alternative Answer –
i) Given that voltage of the battery, V = 6V and resistance are R1 = 1Ω. R2 = 2Ω. are connected in series.
Then equivalent resistance is
Req = R1 + R2
= 1+2
Req = 3Ω
Using Ohm’s law,
V = IReq
I = V/ Req
= 6/3
I = 2A
w.k.t Power, P = I2R
= 22 × 2
= 4 × 2
P = 8W
The power used in the 2Ω resistore of the circuit in a 6V battery in series with 1Ω and 2Ω resistores is 8 W
ii) Given that voltage of the battery, V = 4V and resistores are R1 = 12Ω and R2 = 2Ω connected in parallel combination.
When resistors are connected in parallel then the voltage throughout the circuit is same.
Using Joule’s law of heating
Power, P = V2/ R
= 42/2
= 16 / 2
P = 8W
The power used in the 2Ω resistore in a 4V battery in parallel with 12Ω and 2Ω resistores is 8W.
Therefore in both cases power is used 8W.
15.) Two lamps, one rated 100 W at 220V, and the other 60W at 220V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Ans:
Given that,
- One lamp rated 100W at 220V and second lamp rated 60W at 220V are connected in parallel then potential difference across each lamp in parallel combination will be same and which is 220V but current get divided in parallel combination.
- So current drawn by lamp one,
P= VI
- Hence, I= P/V = 100/220= 0.4545 A
- And the current drawn by lamp second will be,
P= VI
Hence, I= P/V = 60/220 = 0.2727 A
- Thus, total current drawn from line by these two lamps connected in parallel will be,
I= 0.4545 + 0.2727= 0.7272 A
Alternative Answer –
Given that,
Power of the first lamp, P1 = 100W
And voltage, V1 = 220 V
Power of the second lamp, P2 = 60W
And voltage, V2 = 220 V
To find resistance of the first lamp,
We use Joule’s law of heating,
P = V2 / R
R = V2 / P
Therefore R1 = ( V1 )2 / P1
R1 =( 220)2 /100
To find resistance of the second lamp
We use the formula
R2 =( V2 )2 / P2
R2 = (220)2/60
Given that both lamps are connected in parallel so,
1/Req = (1/R1) + (1/R2)
1/ Req = (1/( (220)2/100))) + (1/((220)2/60))
1/ Req = (100+60)/(220)2
Req = (220)2/160
To find current we use Ohm’s law,
V = IReq
I = V/Req
I = 220/((220)2/160)
I = 160/220
I = 0.727A
Therefore 0.727A current is drawn from the line if the supply voltage is 220 V
16.) Which uses more energy, a 250W TV set in 1hr or a 1200 W toaster in 10 minutes?
Ans:
- We have to find the energy consumed by a TV set of 250W in 1hr.
P= 250W
time = 60 min = 60*60= 3600 second
- So energy consumed by TV set will be,
E= P*t = 250*3600= 900,000 J = 900 KJ
- Now, we have to find the energy consumed by a toaster of 1200W in 10 minutes.
P= 1200 W
Time = 10 min = 10*60= 600 second
- Thus, energy consumed by a toaster of 1200 W in 10 min will be,
E= P*t = 1200*600 = 720,000J = 720KJ
- Thus, we can say that a TV set of 250 W uses more energy in 1 hr than the toaster of 1200 W in 10 minutes.
Alternative Answer –
Given that,
The power of the TV, P1 = 250 W
The time for the TV, t1 = 1 hr
The power of the Toaster, P2 = 1200 W
The time for Toaster, t2 = 10 mins = (10/60) hr
w.k.t the formula, E = Pt
Where E is energy,P is power and t is the time
- Energy used by Tv, E1 = P1t1
E1 = 250 × 1
E1 = 250 W hr
- Energy used by Toaster, E2 = P2t2
= 1200× (10/60)
= 200 W hr
Therefore Tv used more energy than the Toaster.
17.) An electric heater of resistance 8 ohm draws 15A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater?
Ans:
Given that,
Resistance of heater R= 8 ohm
Current drawn by heater I= 15A
Time = 2 hrs
- We know that, power is given by,
P= I2*R = 225*8 = 1800 W = 1800 J/s
- Thus, the electric heater of resistance 8 ohm drawing current 15 A from the service mains 2 hours, generates heat at the rate of 1800 J /s.
Alternative Answer –
Given that,
The resistance of the heater, R = 8Ω
Current, I = 15A
Using the formula, P = VI
= (IR)I
= I2 R
= (15)2× 8
= 225 × 8
= 1800 W
Therefore the rate at which heat is developed in the heater is 1800 W.
18.) Explain the following:
a) why is the tungsten used almost exclusively for filament of electric lamps?
Ans:
We know that, if electric lamps are on then they get heated as time goes due to which high temperature will be created.
Tungsten is an alloy which is having high melting point and high resistivity due to which it does not melt even at high temperature. Because of this reason tungsten is used almost exclusively for filament of electric lamps.
Alternative Answer –
- Tungsten has very high tensile strength compared to other metals. To emit light in an electric bulb,we need a metal that, when heated to a temperature that emits the most light,is tungsten. Tungsten has a very high melting point (3380°C ) and has a low vapour pressure. Tungsten is used in electric bills because it emits excessive light when it reaches high temperature before meeting.
b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
Ans: The conductors of heating devices such as bread-toasters and electric irons are generating more heat and heat generated is directly proportional to the resistivity of the material used.
And we know that, alloys has more resistivity than the pure metals. Hence, alloys are used rather than pure metal in conductors of electric heating devices.
Alternative Answer –
Compared to impure metals, pure metals are corrode faster when exposed to the air. Also ,pure metals do not have high temperature resistance compared to impure metals,so they melt quickly at high temperature. If we made electric equipment in pure metal then the electrical temperature would melt the pure metals and cause some kind of problems. Therefore Electrical equipments are made from impure metals.
c) Why is the series arrangement not used for domestic circuits?
Ans:
- We know that, in series arrangement voltage gets divided across the instruments used due to which voltage drop across each electric appliance will be less. And hence less current will be drawn by the electric appliances due to which they may get more heated and sometimes appliances may not work properly.
- Because of these all-reasons electric appliances are not connected in series in domestic circuits.
Alternative Answer –
When we connect components in series connection, then electricity passes through each component. In this process, if one component becomes inactive, the electrical connection does not continue and all the next components do not work,and the circuit will break down. But when we connect components in parallel connection, electricity passes through individual components, in this case, even if one component becomes inactive, electricity reaches the next components and continues to work,and the circuit will be complete. So series arrangements are not used for domestic circuits.
d) How does the resistance of a wire vary with its area of cross section?
Ans:
- We know that,
R = qL/A
- Hence, resistance of a wire is inversely proportional to the area of cross section of the wire.
- If area of cross section is increased then resistance of the wire decreases and when area of cross section of the wire is decreased then resistance of the wire increases.
Alternative Answer –
W.k.t R∝ (L/A)
The above reaction shows that resistance is inversely proportional to area of the cross section of wire, therefore from above relation we can say that when resistance of the wire is increases then the Area of the cross section of the wire is decreases,if area of cross section of wire is increases then resistance of the wire decreases.
e) Why are copper and aluminium wires usually employed for electricity transmission?
Ans:
- The materials used in electricity transmission has to consume less power and has to give more power to the electrical equipment.
- And we know that, power consumed is directly proportional to resistivity of the material used.
- As the resistivity of copper and aluminium is very low hence, they don’t consume more power and most of the power is transmitted through them to the electrical equipment operated.
- Because of these reasons copper and aluminium wires usually employed for electricity transmission.
Alternative Answer –
- k.t aluminium and copper metals are very good conductors therefore in the case of transfer of electricity, from this metal,we easily transfer electricity. And it has low resistance power which prevents the loss of electric power from heat ,during transfer of electricity.
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