# NCERT Solutions Class 10 for Science Chapter 12 Electricity

NCERT Solutions Class 10 for **Science Chapter 12 Electricity** : In this post, we will share with you all the detailed **NCERT Solutions of Class 10 Science Chapter 12 Electricity**. This will contain both in-text and back-exercise questions for Science and Social Science, and all exercise questions for Mathematics. For all school and board level examinations, doing all the NCERT Questions is a must.

# Why are NCERT Questions Important?

NCERT Questions and Answers not only help you get hold of concepts firmly and enhance your understanding, but also form the base of all types of questions asked in exams. Questions asked in exam are more or less the same type as mentioned in NCERT. Moreover, sometimes the questions in NCERT are directly asked in exams, as it is, without any changes.

Hence, it’s very important to understand NCERT Questions and Answers.

- Get NCERT Solutions of all Science Chapters of Class 10
- Get NCERT Solutions of all Mathematics Chapters of Class 10
- Get NCERT Solutions of all Social Science Chapters of Class 10
- Get Notes, PYQs, Free E-Books, and other material for Class 10
- Join our Telegram Channel for Latest Updates here

[adinserter block=”3″]

# Class 10 Science Chapter 12 Electricity

[twy_show_add_bookmark]

**In-Text Questions** (Page 200)

**Question 1:**

What does an electric circuit mean ?

Answer 1:

A continuous and closed path along which an electric current flows is called an electric circuit.

**Question 2:**

Define the unit of current.

**Answer 2:**

The unit of current is ampere. Ampere is defined by the flow of one coulomb of charge per second.

**Question 3:**

Calculate the number of electrons constituting one coulomb of charge.

**Answer 3:**

Charge on one electron, e = 1.6 x 10-19 C

Total charge, Q = 1 C

Number of electrons, n = Qe = 1C1.6×10−19 = 6.25 x 1018

[adinserter block=”3″]

## In-Text Questions (Page 202)

**Question 1:**

Name a device that helps to maintain a potential difference across a conductor.

**Answer 1:**

Battery consisting of one or more electric cells is one of the devices that help to maintain a potential difference across a conductor.

**Question 2:**

What is meant by saying that the potential difference between two points is 1 V?

**Answer 2:**

When 1 J of work is done to move a charge of 1 C from one point to another, it is said that the potential difference between two points is 1 V.

**Question 3:**

How much energy is given to each coulomb of charge passing through a 6 V battery?

**Answer 3:**

We know that the potential difference between two points is given by the equation,

V = W/Q, where,

W is the work done in moving the charge from one point to another

Q is the charge

From the above equation, we can find the energy given to each coulomb as follows:

W = V × Q

Substituting the values in the equation, we get

W = 6V × 1C = 6 J

Hence, 6 J of energy is given to each coulomb of charge passing through a 6 V of battery.

[adinserter block=”3″]

## In-Text Questions (Page 209)

**Question 1:**

On what factors does the resistance of a conductor depend?

**Answer 1:**

The resistance of a conductor depends (i) on its length (ii) on its area of cross-section and (iii) on the nature of its material.

**Question 2:**

Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source ? Why ?

**Answer 2:**

The current will flow more easily through a thick wire than a thin wire of the same material. Larger the area of cross-section of a conductor, more is the ease with which the electrons can move through the conductor. Therefore, smaller is the resistance of the conductor.

**Question 3:**

Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it ?

**Answer 3:**

The change in the current flowing through the electrical component can be determined by Ohm’s Law.

According to Ohm’s Law, the current is given by

I = V/R

Now, the potential difference is reduced to half keeping the resistance constant,

Let the new voltage be V’ = V/2

Let the new resistance be R’ = R and the new amount of current be I’.

The change in the current can be determined using Ohm’s law as follows:

Therefore, the current flowing the electrical component is reduced by half.

**Question 4:**

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

**Answer 4:**

The melting point of an alloy is much higher than a pure metal because of its high resistivity. At high temperatures, alloys do not melt readily. Therefore, alloys are used in heating appliances such as electric toasters and electric irons.

**Question 5:**

Use the data in Table 12.2 (in NCERT Book on Page No. 207) to answer the following :

(i) Which among iron and mercury is a better conductor ?

(ii) Which material is the best conductor ?

**Answer 5:**

(i) Resistivity of iron = 10.0 x 10-8 Ω m

Resistivity of mercury = 94.0 x 10-8 Ω m.

Thus iron is a better conductor because it has lower resistivity than mercury.

(ii) Because silver has the lowest resistivity (= 1.60 x 10-8 Ω m), therefore silver is the best conductor.

[adinserter block=”3″]

## In-Text Questions (Page 213)

**Question 1:**

Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

**Answer 1:**

A battery of three cells of 2 V each equals to battery of potential 6 V. The circuit diagram below shows three resistors of resistance 12 Ω, 8 Ω and 5 Ω connected in series along with a battery of potential 6 V.

**Question 2:**

Redraw the circuit of Questions 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter ?

**Answer 2:**

An ammeter should always be connected in series with resistors while the voltmeter should be connected in parallel to the resistor to measure the potential difference as shown in the figure below.

Using Ohm’s Law, we can obtain the reading of the ammeter and the voltmeter.

The total resistance of the circuit is 5 Ω + 8 Ω +12 Ω = 25 Ω.

We know that the potential difference of the circuit is 6 V, hence the current flowing through the circuit or the resistors can be calculated as follows:

I = V/R = 6/25 = 0.24A

Let the potential difference across the 12 Ω resistor be V1.

From the obtained current V1 can be calculated as follows:

V1 = 0.24A × 12 Ω = 2.88 V

Therefore, the ammeter reading will be 0.24 A and the voltmeter reading be 2.88 V.

## In-Text Questions (Page 216)

**Question 1:**

Judge the equivalent resistance when the following are connected in parallel :

(i) 1 Ω and 106 Ω,

(if) 1 Ω and 103 Ω and 106 Ω.

**Answer 1:**

When the resistances are connected in parallel, the equivalent resistance is smaller than the smallest individual resistance.

(i) Equivalent resistance < 1 Ω.

(ii) Equivalent resistance < 1 Ω.

**Question 2:**

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω 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 ?

**Answer 2:**

Resistance of electric lamp, R1 = 100 Ω

Resistance of toaster, R2 = 50 Ω

Resistance of water filter, R3 = 500 Ω

Equivalent resistance Rp of the three appliances connected in parallel, is

Resistance of electric iron = Equivalent resistance of the three appliances connected in parallel = 31.25 Ω

Applied voltage, V = 220 V

Current, I = VR = 220V31.25Ω

**Question 3:**

What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ?

**Answer 3:**

Advantages of connecting electrical devices in parallel with the battery are :

1.In parallel circuits, if an electrical appliance stops working due to some defect, then all other appliances keep working normally.

2.In parallel circuits, each electrical appliance has its own switch due to which it can be turned on turned off independently, without affecting other appliances.

3.In parallel circuits, each electrical appliance gets the same voltage (220 V) as that of the power supply line.

4.n the parallel connection of electrical appliances, the overall resistance of the household circuit is reduced due to which the current from the power supply is high.

**Question 4:**

How can three resistors of resistances 2Ω, 3 Ω, and 6Ω be connected to give a total resistance of (i) 4 Ω, (ii) 1 Ω ?

**Answer 4:**

i) We can get a total resistance of 4Ω by connecting the 2Ω resistance in series with the parallel combination of 3Ω and 6Ω.

(ii) We can obtain a total resistance of 1Ω by connecting resistors of 2 Ω, 3 Ω and 6 Ω in parallel.

**Question 5:**

What is (i) the highest, (ii) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

**Answer 5:**

(a) If the four resistors are connected in series, their total resistance will be the sum of their individual resistances and it will be the highest. The total equivalent resistance of the resistors connected in series will be 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.

(b) If the resistors are connected in parallel, then their equivalent resistances will be the lowest.

Their equivalent resistance connected in parallel is

Hence, the lowest total resistance is 2 Ω.

[adinserter block=”3″]

## In-Text Questions (Page 218)

**Question 1:**

Why does the cord of an electric heater not glow while the heating element does ?

**Answer 1:**

Heat generated in a circuit is given by I2R t. The heating element of an electric heater made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current because of its high resistance, but the cord of the electric heater made of copper does not glow because negligible heat is produced in it by passing current because of its extremely low resistance.

**Question 2:**

Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

**Answer 2:**

The heat generated can be computed by Joule’s law as follows:

H = VIt

where,

V is the voltage, V = 50 V

I is the current

t is the time in seconds, 1 hour = 3600 seconds

The amount of current can be calculated as follows:

**Question 3:**

An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30 s.

**Answer 3:**

Here, R = 20 Ω, i = 5 A, t = 3s

Heat developed, H = I2 R t = 25 x 20 x 30 = 15,000 J = 1.5 x 104 J

## In-Text Questions (Page 220)

**Question 1:**

What determines the rate at which energy is delivered by a current?

**Answer 1:**

Electric power is the rate of consumption of electrical energy by electric appliances. Hence, the rate at which energy is delivered by a current is the power of the appliance.

**Question 2:**

An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

**Answer 2:**

The power of the motor can be calculated by the equation,

P = VI

Substituting the values in the above equation, we get

P = 220 V × 5 A = 1100 W

The energy consumed by the motor can be calculated using the equation,

E = P × T

Substituting the values in the above equation, we get

P = 1100 W × 7200 = 7.92 × 106 J

The power of the motor is 1100 W and the energy consumed by the motor in 2 hours is 7.92 × 106 J.

[adinserter block=”3″]

## Exercise Questions (Page 221)

**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) 125

(b) 15

(c) 5

(d) 25

**Answer 1:**

d) 25

**Question 2:**

Which of the following terms does not represent electrical power in a circuit?

(a) I2R

(b) IR2

(c) VI

(d) v22

**Answer 2:**

a) IR2

**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 3:**

25 W

**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 4:**

c) 1 : 4

**Question 5:**

How is a voltmeter connected in the circuit to measure the potential difference between two points ?

**Answer 5:**

To measure the voltage between any two points, the voltmeter should be connected in parallel between the two points.

[adinserter block=”3″]

**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 6:**

The resistance of the copper wire of length in meters and area of cross-section m2 is given by the formula

**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 :

[adinserter block=”3″]

**Answer 7:**

The graph between V and I for the above data is given below.

The slope of the graph will give the value of resistance.

Let us consider two points P and Q on the graph.

and from P along Y-axis, which meet at point R.

Now, QR = 10.2V – 34V = 6.8V

And PR = 3 – 1 = 2 ampere

Thus, resistance, R = 3.4 Ω

**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 8:**

Here, V = 12 V and I = 2.5 mA = 2.5 x 10-3 A

∴ Resistance, R = VI = 12V2.5×103A = 4,800 Ω = 4.8 x 10-3 Ω

**Question 9:**

A battery of 9V is connected in series with resistors of 0.2 O, 0.3 O, 0.4 Q, 0.5 Q and 12 £1, respectively. How much current would flow through the 12 Q resistor?

**Answer 9:**

Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω

Potential difference, V = 9 V

Current through the series circuit, I = VR = 12V13.4Ω = 0.67 A

∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.

**Question 10:**

How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

**Answer 10:**

Suppose n resistors of 176 Ω are connected in parallel.

[adinserter block=”3″]

**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) 4Ω

**Answer 11:**

Here, R1 = R2 = R3 = 6 Ω.

(i) When we connect R1 in series with the parallel combination of R2 and R3 as shown in Fig. (a).

The equivalent resistance is

(ii) When we connect a series combination of R1 and R2 in parallel with R3, as shown in Fig. (b), the equivalent resistance is

**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 12:**

Here, current, I = 5 A, voltage, V = 220 V

∴ Maxium power, P = I x V = 5 x 220 = 1100W

Required no. of lamps =Max.PowerPowerof1lamp=110010=110

∴ 110 lamps can be connected in parallel.

**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 13:**

**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 resistor

**Answer 14:**

**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 15:**

Since both the bulbs are connected in parallel, the voltage across each of them will be the same.

Current drawn by the bulb of rating 100 W can be calculated as follows:

P = V × I

I = P/V

Substituting the values in the equation, we get\

I = 100 W/220 V = 100/220 A

Similarly, the current drawn by the bulb of rating 60 W can be calculated as follows:

I = 60 W/220 V = 60/220 A

Therefore, the current drawn from the line is

100/220+60/220=0.727A

**Question 16:**

Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

**Answer 16:**

The energy consumed by electrical appliances is given by the equation

H = Pt, where P is the power of the appliance and t is the time

Using this formula, the energy consumed by a TV of power ration 250 W, can be calculated as follows:

H = 250 W × 3600 seconds = 9 × 105 J

Similarly, the energy consumed by a toaster of power rating 1200 W is

H = 1200 W × 600 s = 7.2 × 105 J

From the calculations, it can be said that the energy consumed by the TV is greater than the toaster.

**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 17**:

Here, R = 8 Ω, 1 = 15 A, t = 2 h

The rate at which heat is developed in the heater is equal to the power.

Therefore, P = I2 R = (15)2 x 8 = 1800 Js-1

**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 copper and aluminium wires are usually employed for electricity transmission?

**Answer 18**:

a. The resistivity and melting point of tungsten is very high. Due to this property, it doesn’t burn readily when heated. Electric lamps operate at high temperature. Hence, tungsten is a choice of metal for the filament of electric lamps.

b. The conductors of electric heating devices are alloys because of their high resistivity. Due to its high resistivity it produces large amount of heat.

c. The voltage is divided in series circuit as result each component in the circuit receives a small voltage because of which the amount of current decreases and the device gets hot and does not work properly. This is the reason why series circuits are not used in domestic circuits.

d. Resistance is inversely proportional to the area of cross section. When the area of cross section increases the resistance decreases and vice versa.

e. Copper and aluminium are good conductors of electricity and have low resistivity because of which they are usually employed for electricity transmission.

[twy_show_add_bookmark]

[adinserter block=”3″]

## For More Content related to Class 10 –

- Get NCERT Solutions of all Science Chapters of Class 10
- Get NCERT Solutions of all Mathematics Chapters of Class 10
- Get NCERT Solutions of all Social Science Chapters of Class 10
- Get Notes, PYQs, Free E-Books, and other material for Class 10
- Join our Telegram Channel for Latest Updates here

[adinserter block=”3″]