# Class 10 Science Chapter 12 Electricity NCERT Solution

Chapter 12 of Science Subject of Class 10 is “Electricity” .NCERT Solutions (Intext and End Exercise Questions) for Science Chapter 12 are provided on this page.

Contents

## Class 10 Science Chapter 12 – Electricity

Chapter 12 of Class 10 Science NCERT Text book is “Electricity”. The end exercise of this chapter consists of a total of 18 questions. These end exercise questions are related to Electricity , 1 ampere, electrons, etc.

## Chapter 12 : Electricity ( NCERT Solutions )

Class 10 Science Chapter 12 Question 1 :
A piece of wire 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

Answer and Explanation : Option [d] “25” is the correct answer.

Class 10 Science Chapter 12 Question 2 :
Which of the following terms does not represent electrical power in a circuit?

(a) I2R
(b) IR2
(c) VI
(d) V2/R

Answer and Explanation : Option [b] “IR2 is the correct answer.

Class 10 Science Chapter 12 Question 3 :
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –

(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W

Answer and Explanation : Option [d] “25 W” is the correct answer.

Question 4 :
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –

(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1

Answer and Explanation : Option [c] “1:4 ” is the correct answer.

Class 10 Science Chapter 12 Question 5 :
How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer and Explanation : A voltmeter is connected in parallel in the circuit to measure the potential difference.

Class 10 Science Chapter 12 Question 6 :
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Answer and Explanation :
Diameter of Copper Wire : 0.5 mm = 0.5/1000 m
Length of wire (L) = ?
Resistance = 10 Ω
Resistivity = 1.6 * 10-8 m
We know the resistance formula
R = ρ L/A
A = π r2
=3.14 * (0.5/2000)2
= 1.9625 * 10-8 m2

Putting all value in Resistance formula
R = ρ L/A
L = R A / ρ
L = 10 *1.9628 *10-8 / (1.6 * 10-8)
L = 122.65 m

If diameter is doubled i.e. 2* 0.5 mm = 0.001 m
Now R = ρ L/A
R = 122.65 m

Class 10 Science Chapter 12 Question 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 the resistance of that resistor.

Answer and Explanation :
Resistance from Graph R = ( V2 – V1 )/( I2 – I1)
R = (10.2 – 6.7 ) / ( 3.0 -2.0 ) = 3.5

Class 10 Science Chapter 12 Question 8 :
When a 12 V 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.

Answer and Explanation :
We know that R = V/I
There fore , R = 12 / ( 2.5 * 10-3 ) = 4.8 * 103 Question 9 :
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Answer and Explanation :
I = V/R
I = 9 / 12 = 0.67 A
Therefore , 0.67 A of current flows through 12 resistor.

Class 10 Science Chapter 12 Question 10 :
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer and Explanation :
Let the number of resistor be n.
Therefore ,
1/R = 1/176 +/176 + 1/176 + ….. + n times
R = 176 /n
Now,
I = V/R
5 = 220 /( 176/n)
n= 4
Number of resistor n = 4

Class 10 Science Chapter 12 Question 11 :
Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of

(i) 9 Ω,
(ii) 4 Ω

Answer and Explanation :
(i) Connect two resistor of 6 ohm in parallel then connect third 6 ohm resistor in series with the parallel combination. This will result an equivalent 9 ohm resistance.
(ii) Connect two resistor in series then connect third one in parallel with the series combination . It will result an equivalent 4 ohm resistance.

Class 10 Science Chapter 12 Question 12 :
Several electric bulbs designed to be used on a 220 V 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 5 A?

Answer and Explanation :
Resistance of each bulb = R
R = V2/P = 220*220 / 10 = 4840 ohm
Let x bulb be connected in parallel to 220 V to obtain max 5A current .
Therefore ,
1/Rp =1/4840 + …… n times
Rp = 4840/x
Now;
I = V/R
5 = 220/(4840 /x )
x = 200 bulbs

Class 10 Science Chapter 12 Question 13 :
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer and Explanation :
Case I : When the plates are connected in Series
R = 24 +24 = 48 ohm
Therefore, I = V/R = 220/48 = 48 A

Case II : When the plate is connected in parallel
1/R = 1/24 +1/24 = 1/12
R = 12 ohm
Therefore, I = V/R = 220/12 = 18.3 A

Case III : When only resistance is connected
I = V/R = 220/24 = 9.2 A

Class 10 Science Chapter 12 Question 14 :
Compare 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, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors

Answer and Explanation :
(i) V = 6V
R= 1 ohm and 2 ohm
Rs = 1 + 2 = 3 ohm
I =V/R = 6/3 = 2A
Therefore, P1 = VI = 6*2 = 12W

(ii) V= 4V
R =12 ohm and 2 ohm
I = V/R = 4/2 = 2 A
P2 = V I = 4 * 2 =8 W

P1 : P2 = 3: 2

Class 10 Science Chapter 12 Question 15 :
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Answer and Explanation :
Lamp (1) = P = 100 W
I1 = P/V = 100 /220 = 10/22 A
Lamp (2) = P = 60 W
I2 = P/V = 60/220 = 6/22 A
Total Current drawn from line is I1 + I2 = 10/22 +6/22 = 8/11 A
Total Current drawn from line is 0.727 Ampere.

Class 10 Science Chapter 12 Question 16 :
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer and Explanation :
Energy 1 = Power * time
= 250 * 60 * 60
= 15000 * 60 = 900 KJ
Energy 2 = P * t
= 1200 *60 *60
= 720000 =720 KJ
E2 > E1 .
T.V consumes more energy.

Class 10 Science Chapter 12 Question 17 :
An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Answer and Explanation :
Rate of heat = Power = I2Rt/t
= I2R = 15 *15 * 8 = 1800 Watt
So, heat is developed at the rate of 1800 J/sec

Class 10 Science Chapter 12 Question 18 :
Explain the following.

(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(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?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?

Answer and Explanation :
(a) Tungsten is used almost exclusively for filament of electric lamps because it has high melting point and high resistance.
(b) Conductors of electric heating devices are made of alloy because it has high melting point, high resistance. It does not burn at high temperature.
(c) Series arrangement is not used for domestic circuits because in series arrangement if one of the device stops working other device will also get affected.
(d) Resistance is inversely proportional to area of cross section.
(e) Silver , Copper , aluminium are the best conductor of electricity . Since Silver is too costly , copper and aluminium are used for electricity transmission.