# NCERT Solutions for Class 10 Science Chapter 12 Electricity

NCERT solutions for class 10 science chapter 12 Electricity: While doing preparation for class 10 board exams, you must know that chapter 12 Electricity is one of the most important chapters. This is a scoring chapter and also very useful when you go to class 12. So it will be good for you if you clear your concepts in class 10 itself. The solutions of NCERT for class 10 science chapter 12 Electricity will help you understand that electrical energy is one of the major forms of energy that we are using in our daily lives. One of the important features of electric energy is that it can be transmitted to a large distance with very little power loss. The electrical energy is obtained from various sources such as water, steam, and nuclear energy, etc.

In the CBSE NCERT solutions for class 10 science chapter, 12 Electricity questions related to direct current are discussed. Direct current is unidirectional current whereas the alternating current changes its direction periodically. The measurement of the current is done with the help of ammeter and voltage with the help of a voltmeter. Another important topic that you can understand with the help of solutions for NCERT class 10 science chapter 12 Electricity is a combination of resistors in series and parallel. The main aim of NCERT solutions for class 10 science chapter 12 Electricity is to give a better knowledge of how to use the concept while answering the questions. The NCERT solutions will help you score well in the CBSE board exam.

The solutions of NCERT class 10 science chapter 12 electricity deal with the questions related to the concept of:

• Electric current potential
• Electric resistance
• Circuit diagrams
• Ohm's law
• The heating effect of electric current and power.
• The electric current can be classified as direct current (DC) and alternating current ac (AC).

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Certain points to remember from solutions for class 10 science chapter 12 electricity for NCERT

• Current always flows from positive to the negative terminal of a circuit
• An ammeter is always connected in series with the circuit whose current is to be measured
• a voltmeter is always connected in parallel across a circuit whose voltage is to be measured
• Reciprocal of the resistance is known as conductance
• Reciprocal of resistivity is known as conductivity

## NCERT solutions for class 10 science chapter 12 Electricity

Topic 12.1 Electric Current and Circuit

A closed and continuous path of electric current is known as the electric circuit. It consists of electric circuit elements like battery, resistors etc and electric devices like a switch and measuring devices like ammeters etc.

Ampere(A) is unit of current.1A is flow of 1C of charge through a wire in 1s of time.

$\\current\ I=\frac{q}{t}\\1A=\frac{1 \ C}{1\ sec}$

Given: Q=1C

We know that the charge of an electron $e=1.6\times 10^{-19}C$

$Q=ne$

$\Rightarrow n=\frac{Q}{e}$

$\Rightarrow n=\frac{1}{1.6\times 10^{-19}}=6.25\times 10^{18}$

Thus,  the number of electrons constituting one coulomb of charge is $6\times 10^{18}$ electrons.

## NCERT free solutions for class 10 science chapter 12 Electricity

Topic 12.2 Electric Potential and Potential Difference

Battery, cell or power supply source helps to maintain a potential difference across a conductor.

The potential difference between two points is 1 V means 1 J of work is required to move a charge of amount 1C from one point to another.

Given : potential difference = 6V and carge =1C.

$Potential\, \, difference=\frac{work\, \, done}{charge}$

$\Rightarrow 6=\frac{work\, \, done}{1}$

$\Rightarrow work\, \, done=6\times 1=6$J

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

## Solutions for NCERT class 10 science chapter 12 Electricity

Topic 12.5 Factors on which the resistance of a conductor depends

The resistance of a conductor depends on :

1. Cross section area of the conductor.

2.length of conductor

3.The temperature of the conductor.

4. Nature of material of the conductor.

We know that resistance is given as

$R=\frac{\rho l}{A}$

R=resistance

$\rho$=resistivity

l=length of wire

A=area of cross section

Resistance is inversely proportional to the area of cross-section.

Thicker the wire more is cross-sectional area resulting in less resistance resulting in more current flow.

By Ohm's law,

V=IR

$I=\frac{V}{R}$

V=  potential difference

I =current

R = resistance

Now, the potential difference is reduced to half i.e.

$V'=\frac{V}{2}$

R' = R=resistance

I' =current

$I'=\frac{\frac{V}{2}}{R}=\frac{V}{2R}=\frac{I}{2}$

Current flowing is reduced to half.

The resistivity of an alloy is higher than pure metal so alloy does not melt at high temperature. Thus, coils of electric toasters and electric irons made of an alloy rather than a pure metal.

(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?

Lower the value of resistivity lower will be the resistance for a material of given area and length. If resistance is law current flow will be high when a potential difference is applied across the conductor.

(a).resistivity of iron =$10.0\times 10^{-8} \Omega m$

the resistivity of mercury =$94\times 10^{-8} \Omega m$

the resistivity of mercury is more than iron so iron is a better conductor than mercury.

(b).From the table, we can observe silver has the lowest resistivity so it is the best conductor.

## NCERT solutions for class 10 science chapter 12 Electricity

Topic 12.6 Resistance of a system of resistors

The schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a $5 \Omega$ resistor, an $8 \Omega$ resistor, and a $12 \Omega$ resistor, and a plug key, all connected in series is as shown below :

The diagram is as shown :

Resistance of circuit =R=5+8+12=25

Potential = 6V

$V=IR$

$\Rightarrow I=\frac{V}{R}=\frac{6}{25}=0.24A$

Now, for 12 ohm resistor, current = 0.24 A.

By Ohm's law,

$V=IR=0.24\times 12=2.88V$

The reading of the ammeter is o.24A and the voltmeter is 2.88V.

## Class 10 science chapter 12 Electricity solutions for NCERT

Topic 12.6.2 Resistors in Parallel

$(a)\; 1\; \Omega \; and \; 10^{6}\Omega$        $(b)\; 1\; \Omega \; and \; 10^{3}\Omega,\; and \; 10^{6}\Omega,$

$(a)\; 1\; \Omega \; and \; 10^{6}\Omega$

R=Equivalent  resistance

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\Rightarrow \frac{1}{R}=\frac{1}{1}+\frac{1}{10^{6}}$

$\Rightarrow \frac{1}{R}=\frac{10^6+1}{10^{6}}$

$\Rightarrow R=\frac{10^6}{10^{6}+1}\approx 1\Omega$

$(b)\; 1\; \Omega \; and \; 10^{3}\Omega,\; and \; 10^{6}\Omega,$

R=Equivalent  resistance

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\Rightarrow \frac{1}{R}=\frac{1}{1}+\frac{1}{10^3}+\frac{1}{10^{6}}$

$\Rightarrow \frac{1}{R}=\frac{10^6+10^3+1}{10^{6}}$

$\Rightarrow R=\frac{10^6}{10^{6}+10^3+1}=0.999\Omega \approx 1\Omega$

Given : $R_1=100\Omega ,R_2=50\Omega ,R_3=500\Omega$

R=Equivalent  resistance

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\Rightarrow \frac{1}{R}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}$

$\Rightarrow \frac{1}{R}=\frac{5+10+1}{500}=\frac{16}{500}$

$\Rightarrow R=\frac{500}{16}=31.25\Omega$

By Ohm's law,

$I=\frac{V}{R}=\frac{220}{31.25}=7.04A$

Hence, the resistance of electric iron is 31.25 and current through it is 7.04A.

In parallel, there is no division of voltage among the appliances so the potential difference across all appliance is equal and the total effective resistance of a circuit can be reduced. Hence, connecting electrical devices in parallel with the battery is beneficial instead of connecting them in series.

$(a)4\Omega , (b)1\Omega ?$

(a) $R_1=3,R_2=6,R_3=2$

R=Equivalent  resistance

$\frac{1}{R_1_2}=\frac{1}{R_1}+\frac{1}{R_2}$

$\Rightarrow \frac{1}{R_1_2}=\frac{1}{3}+\frac{1}{6}$

$\Rightarrow \frac{1}{R_1_2}=\frac{2+1}{6}$

$\Rightarrow R_1_2=\frac{6}{3}=2$

$R=R_1_2+R_3=2+2=4$

(b).1 Ohm

R=Equivalent  resistance

Connect all the three resistors in parallel

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\Rightarrow \frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$

$\Rightarrow \frac{1}{R}=\frac{6+3+1}{6}$

$\Rightarrow R=\frac{16}{6}=1\Omega$

•   $4\; \Omega , 8\; \Omega ,12\; \Omega ,24\; \Omega ?$

(a) the highest total resistance is obtained when all the resistors are connected in series

$4+8+12+24=48\Omega$

• $4\; \Omega , 8\; \Omega ,12\; \Omega ,24\; \Omega ?$

(b) the lowest total resistance is R is obtained when all the resistors are connected in parallel.

$\frac{1}{R}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}$

$\frac{1}{R}=\frac{6+3+2+1}{24}=\frac{12}{24}$

$R=\frac{24}{12}=2$

## NCERT free solutions for class 10 science chapter 12 Electricity

The heating element of an electric heater is a resistor.

The amount of heat production is given as ,   $H=I^2Rt$.

The resistance of elements (alloys) of an electric heater is high. As current flow through this element, it becomes hot and glows red.

The resistance of cord (metal like Cu or Al)of an electric heater is low. As current flow through this element, it does not becomes hot and does not glow red.

Given : potential  difference = 50 V.

Charge = 9600 C

time = 1 hr=3600 s

$I=\frac{charge}{time(seconds)}=\frac{9600}{3600}=\frac{80}{3}A$

So,     $H=VIt$

$\Rightarrow H=50\times \frac{80}{3}\times 3600$

$\Rightarrow H=4800000J=4.8\times 10^6J$

Given :  resistance=R= $20 \; \Omega$, TIME =30s    , current =I=5A

$V=IR$

$\Rightarrow V=5\times 20=100V$

$H=VIt$

$\Rightarrow H=100\times 5\times 30$

$\Rightarrow H=15000J=1.5\times 10^4J$

## NCERT textbook solutions for class 10 science chapter 12 Electricity

Topic 12.8 Electric power

## Q.1  What determines the rate at which energy is delivered by a current?

The rate at which energy is delivered by a current is power.

Power P=I2R, where I is the current and R is the resistance of the circuit/ appliance. Power depends on the current drawn by the appliance and the resistance of the appliance.

Given: I= 5 A  and   V= 220 V.

Power=P=VI

$P=220\times 5=1100W$

Time = 2hr=$2\times 60\times 60=7200s$

The energy consumed = $power\times time$

$=1100\times 7200J=7920000J$

Power of motor = 1100W

Energy consumed by motor = 7920000J

NCERT solutions for class 10 science chapter 12 Electricity- Exercise solutions

(a) 1/25             (b) 1/5             (c) 5             (d) 25

Given: A piece of wire of resistance $R$ is cut into five equal parts.

Resistance of each part is $\frac{R}{5}$.

$\frac{1}{R'}=\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}$

$\frac{1}{R'}=\frac{25}{R}$

$R'=\frac{R}{25}$

Hence, $\frac{R}{R'}=\frac{R}{\frac{R}{25}}=25$

Thus, option D is correct.

$(a)\; I^{2}R$                $(b)\; IR^{2}$            $(c)\; VI$               $(d)\; V^{2}/R$

We know that power = P=VI....................................1

Put , V=IR in equation 1

$P=I^2R$

Pur , $I=\frac{V}{R}$    in equation 1,

$P=V\times \frac{V}{R}=\frac{V^2}{R}$

P= power, V=potencial difference, I= current,R=resistance

P cannot be $IR^{2}$.

Thus, option B is correct.

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

Given : V=220V,P=100W

$P=\frac{V^2}{R}$

The resistance of the bulb

$\Rightarrow R=\frac{V^2}{P}=\frac{(220)^2}{100}=484\Omega$

If bulb is operated on 110 Vand resistance is the same , the power consumed will be P'

$P'=\frac{V^2}{R}=\frac{(110)^2}{484}=25W$

Hence, option D is correct.

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

If resistors are connected in parallel, the net resistance is given as

$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}$

$\Rightarrow \frac{1}{R_p}=\frac{1}{R}+\frac{1}{R}$

$\Rightarrow \frac{1}{R_p}=\frac{2}{R}$

$\Rightarrow R_p=\frac{R}{2}$

If resistors are connected in series, the net resistance is given as

$\Rightarrow R_s=R_1+R_2=R+R=2R$

Heat produced = H = $\frac{V^2}{R}t$

$\frac{H_s}{H_p}=\frac{\frac{V^2t}{2R}}{\frac{V^2t}{\frac{R}{2}}}$

$\Rightarrow \frac{H_s}{H_p}=\frac{1}{4}$

$\Rightarrow H_s:H_p=1:4$

Thus, option C is correct.

A  voltmeter should be connected in parallel,  to measure the potential difference between two points.

Given : diameter=d= 0.5 mm and resistivity = $\rho$=$1.6\times 10^{-8}\; \Omega$ m.,resistance =R=10

Area =A

$A=\frac{\pi d^2}{4}=\frac{3.14\times 0.5\times 0.5}{4}$

$\Rightarrow A=0.000000019625 m^2$

We know

$R=\frac{\rho l}{A}$

$\Rightarrow l=\frac{RA}{\rho }=\frac{10\times 0.000000019625}{1.6\times 10^-^8}$

$=122.72m$

If the diameter is doubled.

d=1 mm

Area =A'

$A'=\frac{\pi d^2}{4}=\frac{3.14\times 1\times 1}{4}$

$\Rightarrow A=0.000000785 m^2$

We know

$R'=\frac{\rho l}{A'}$

$\Rightarrow R'=\frac{1.6\times 10^{-8}\times 122.72}{0.000000785}$

$\Rightarrow R'=2.5\Omega$

$\frac{R'}{R}=\frac{2.5}{10}=\frac{1}{4}$

Hence, new resistance is  $\frac{1}{4}$  of original resistance.

I (amperes)     0.5     1.0     2.0     3.0     4.0
V (volts)          1.6      3.4    6.7     10.2   13.2

Plot a graph between V and I and calculate the resistance of that resistor

The plot between voltage and current is as shown :

The slope of the line gives resistance (R)

$R=\frac{6.8}{2}=3.4\Omega$

Given : V=12V  , I = 2.5 mA = 0.0025 A

$R=\frac{12}{0.0025}=4800\Omega =4.8k\Omega$

Total resistance  =R

$R=0.2+0.3+0.4+0.5+12=13.4\Omega$

V = 9V

$I=\frac{V}{R}=\frac{9}{13.4}=0.67 A$

Hence, All resistors are in series so 0.67A current would flow through the $12\Omega$  resistor.

Given: V=220V and I =5A

$R=\frac{V}{I}=\frac{220}{5}=44\Omega$

Let x number of resistors are connected in parallel to obtain 44 Ohm equivalent resistance.

Thus,

$\frac{1}{R}=\frac{1}{176}+\frac{1}{176}+.......................to\, \, x\, \, times$

$\Rightarrow \frac{1}{44}=\frac{x}{176}$

$\Rightarrow x=\frac{176}{44}=4$

Hence, 4 resistors of 176 Ohm are connected in parallel to obtain 44 Ohm.

(i) $R_1=R_2=R_3=6$

R=Equivalent  resistance

$\frac{1}{R_1_2}=\frac{1}{R_1}+\frac{1}{R_2}$

$\Rightarrow \frac{1}{R_1_2}=\frac{1}{6}+\frac{1}{6}$

$\Rightarrow \frac{1}{R_1_2}=\frac{1+1}{6}$

$\Rightarrow R_1_2=\frac{6}{2}=3$

$R=R_1_2+R_3=3+6=9\Omega$

(ii)

.$R_1=R_2=R_3=6$

R=Equivalent  resistance

$R_1_2=R_1+R_2=6+6=12\Omega$

$\frac{1}{R}=\frac{1}{R_1_2}+\frac{1}{R_3}$

$\Rightarrow \frac{1}{R}=\frac{1}{12}+\frac{1}{6}$

$\Rightarrow \frac{1}{R}=\frac{1+2}{12}$

$\Rightarrow R=\frac{12}{3}=4$

Given: V=220V and P=10W

$R=\frac{V^2}{P}=\frac{220^2}{10}=4840\Omega$

Let x be the number of bulbs.

I = 5A              and      V=220V

$R=\frac{V}{I}=\frac{220}{5}=44\Omega$

For x bulbs of resistance 4680 Ohm, are connected in parallel to obtain 44 Ohm equivalent resistance.

Thus,

$\frac{1}{R}=\frac{1}{4840}+\frac{1}{4840}+.......................to\, \, x\, \, times$

$\Rightarrow \frac{1}{44}=\frac{x}{4840}$

$\Rightarrow x=\frac{4840}{44}=110$

Hence, 110 bulbs of 4840 Ohm are connected in parallel to obtain 44 Ohm.

Given : V=220V and Resistance of each coil=R =24A

When coil is used separately,current in coil is

$I=\frac{V}{R}=\frac{220}{24}=9.16A$

When two coils are connected in series, net resistance is

$R=R_1+R_2=24+24=48\Omega$

current in coil is I'

$I'=\frac{V}{R}=\frac{220}{48}=4.58A$

When two coils are connected in parallel, net resistance is

$\frac{1}{R}=\frac{1}{24}+\frac{1}{24}=\frac{2}{24}$

$\Rightarrow R=12\Omega$

current in the coil is I''

$I''=\frac{V}{R}=\frac{220}{12}=18.33A$

(i) a 6 V battery in series with $1\Omega$ and $2\Omega$  resistors

(ii) a 4 V battery in parallel with $12\Omega$ and $2\Omega$ resistors.

i) Given: V=6V

R=1+2=3Ohm

$I=\frac{V}{R}=\frac{6}{3}=2A$

In, a series current is constant.

So, power =P

$P=I^2R=2^2\times 2=8W$

ii) Given: V=6V , R=2 Ohm

In,  parallel combination voltage in the circuit is constant.

So, power =P

$P=\frac{V^2}{R}=\frac{4^2}{2}=8W$

Given :

For lamp one:  Power = P1=100W    and        V = 220V

$I_1=\frac{P_1}{V}=\frac{100}{220}=0.455A$

For lamp two:  Power = P2=60W    and        V = 220V

$I_2=\frac{P_2}{V}=\frac{60}{220}=0.273A$

Thus, the net current drawn from the supply is 0.455+0.273=0.728A

For TV set :

Given :  Power = 250W  and  time = 1 hr = 3600 seconds

Energy consumed = H=PI

$H=250\times 3600=900000J$

For toaster :

Given :  Power = 1200W  and  time = 10 minutes = 600 seconds

Energy consumed = H=PI

$H=1200\times 600=720000J$

Thus, the TV set uses more energy than a toaster.

## Q.17.   An electric heater of resistance $8\Omega$  draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Given: R=8 Ohm ,I=15A  and t= 2hr

The heat developed in the heater is H.

$H=I^2Rt$

The rate at which heat is developed is given as

$\frac{I^2Rt}{t}=I^2R=15^2\times 8=1800J/s$

Q.18(a).    Explain the following.

(a) Why is the tungsten used almost exclusively for filament of electric lamps?

The tungsten used almost exclusively for filament of electric lamps because the melting point of tungsten is very high. The bulb glows at a very high temperature. So tungsten filament does not melt when bulb glows.

Q.18.(b)    Explain the following.

(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?

The conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal because the resistivity of the alloy is more than a pure metal. So the resistance will be hight and the heating effect will be high.

Q.18.(c)    Explain the following.

(c) Why is the series arrangement not used for domestic circuits?

If any of the elements in the circuit get damaged the entire circuit will be affected. If an element brake there will not be any current flow through the circuit. So the series circuit is not preferred in the domestic circuit.

Q.18.(d)    Explain the following.

(d) How does the resistance of a wire vary with its area of cross-section?

We know that

$R=\frac{\rho l}{A}$

$R\alpha \frac{1}{A}$

Thus,  the resistance of a wire is inversely proportional to its area of cross-section.

Q.18.(e)    Explain the following.

(e) Why are copper and aluminium wires usually employed for electricity transmission?

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

## NCERT Solutions for Class 10 Science - Chapter wise

 Chapter No. Chapter Name Chapter 1 NCERT free solutions for class 10 science chapter 1 Chemical Reactions and Equations Chapter 2 NCERT solutions for class 10 science chapter 2 Acids, Bases, and Salts Chapter 3 Solutions for NCERT class 10 science chapter 3 Metals and Non-metals Chapter 4 NCERT solutions for class 10 science chapter 4 Carbon and its Compounds Chapter 5 NCERT free solutions for class 10 science chapter 5 Periodic Classification of Elements Chapter 6 CBSE NCERT solutions for class 10 science chapter 6 Life Processes Chapter 7 NCERT solutions for class 10 science chapter 7 Control and Coordination Chapter 8 NCERT free solutions for class 10 science chapter 8 How do Organisms Reproduce? Chapter 9 NCERT textbook solutions for class 10 science chapter 9 Heredity and Evolution Chapter 10 Solutions of NCERT class 10 science chapter 10 Light Reflection and Refraction Chapter 11 NCERT free solutions for class 10 science chapter 11 The Human Eye and The Colorful World Chapter 12 Solutions for CBSE NCERT class 10 science chapter 12 Electricity Chapter 13 NCERT free solutions for class 10 science chapter 13 Magnetic Effects of Electric Current Chapter 14 CBSE NCERT solutions for class 10 science chapter 14 Sources of Energy Chapter 15 NCERT textbook solutions for class 10 science chapter 15 Our Environment Chapter 16 Solutions for NCERT class 10 science chapter 16 Sustainable Management of Natural Resources

## Highlights of Solutions for NCERT class 10 science Chapter 12 Electricity

• The solutions provided for chapter 12 Electricity will boost your knowledge.
• With the help of NCERT solutions for class 10 science chapter 12 Electricity, you will understand the concepts easily.
• While preparing for class 10 board exams, practicing with good study material plays a very important role.
• NCERT solutions for class 10 science is the best study material you can have for exam preparation.