# NCERT Solutions for Class 10 Science Chapter 12 Electricity

NCERT Solutions for Class 10 Science Chapter 12 Electricity: Electrical energy is one of the major forms of energy that we are using in our daily life. One of the important features of electric energy is that it can be transmitted to a large distance with very less power loss. The electrical energy is obtained from various sources such as water, steam and nuclear energy etc. This NCERT Solutions for Class 10 Science Chapter 12 introduces the concept of electric current, potential, electric resistance, circuit diagrams, Ohm's law, the heating effect of electric current and power. This chapter of NCERT Solutions for Class 10 Science also gives an introduction to a few circuit elements like a switch, resistor, rheostat, battery, ammeter, and voltmeter.

The electric current can be classified as direct current (DC) and alternating current ac (AC). In this NCERT solution for Class 10 Science, we will focus on direct current. 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. In this chapter of NCERT Solutions for Class 10, we will study both the ammeter and voltmeter and how they are connected in a circuit. Another important topic that we will go through the NCERT Solutions of this chapter is a combination of resistors in series and parallel. The main topics of NCERT Solutions for Class 10 Science Chapter 12 Electricity are given below.

12.1 Electric Current and Circuit

12.2 Electric Potential and Potential Difference

12.3 Circuit Diagram

12.4 Ohm's Law

12.5 Factors on Which the Resistance of a Conductor Depends

12.6 Resistance of a System of Resistors

12.7 Heating Effect of Electric Current

12.8 Electric Power

The  NCERT Solutions for Class 10 Science Chapter 12 starts with the concept of current. What is the definition of current? Current is the rate of flow of charge. That is if a charge of quantity q flows uniformly through a given area for time t then current through the area is $i=\frac{q}{t}$

The unit of current is Ampere and the unit of charge is Columb.

$1\ Ampere=\frac{1\ Columb}{1\ sec}$

That is one-ampere current is given by the flow of one columb of charge for one second through a given area

What is Potential difference in an electrical circuit?

The potential difference between any two points in a circuit is the work done to move a charge from one point of the circuit to another point. That is

$V=\frac{W}{q}$

where V is the potential difference and W is the work done and q is the charge.

When one Joule of work is done to move one Columb of charge from one point to another point of an electric circuit then the potential difference is 1 volt.

Another important topic of NCERT Solutions for Class 10 Science Chapter 12 Electricity is Ohm's law.

What is Ohm's law?

The current through a metallic wire is directly proportional to the potential difference applied across the wire, provided its temperature remains the same.

That is $V\propto I$  where V is the potential difference and I is the current flowing

or

$V=IR$ where R is the resistance. The unit of resistance is ohm. That is the graph between I and V will be a straight line and the slope of the graph gives the value of resistance.

Resistance is the opposition offered to the flow of current. Factors affecting the resistance are also mentioned in the NCERT Solutions for Class 10 Science Chapter 12 Electricity. The factors affecting the resistance are the resistivity(denoted by $\rho$) of the material, length(l) and area(A) of the material. That is resistance R is related to above-mentioned quantities as follows

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

The next topic of the NCERT Solutions for Class 10 Science Chapter 12 is the combination of resistors in series and parallel. When we combine resistance in series the current through the resistors remains the same and the potential difference gets distributed.

Resistors in series

That is the potential difference V=V1+V2+V3

Since current flowing is the same

$\\V=IR_1+IR_2+IR_3$

If the total resistance of the circuit is R then V=IR, therefore

$\\IR=IR_1+IR_2+IR_3\\or\ R=R_1+R_2+R_3$

The net resistance in series combination is the sum of the resistors.

Resistors in Parallel

The next part of the NCERT Solutions for Class 10 Science Chapter 12 Electricity is a parallel combination of resistors.

In a parallel circuit, the voltage across the resistors remains the same but the current get distributed. That is

$\\I=I_1+I_2+I_3$

If R is the total resistance of the circuit then

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

Therefore

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

From the above equation, we can find the value of R. Can you find out the general expression for resistors in series and parallel?

Next topic of the NCERT Solutions for Class 10 Science Chapter 12  is the heating effect of electric current and power

If a current I is flowing through a resistor R for time t then the heat produced in the resistor R is given by

$H=I^2Rt=\frac{V^2}{R}t$

Where V is the potential across the resistor. The unit of heat produced is Joule

Power:  Power is the rate of change of energy, here power

$P=\frac{H}{t}=I^2R=\frac{V^2}{R}$

The unit of power is watt.

Let's solve questions based on NCERT Solutions for Class 10 Science Chapter 12 Electricity

Q1) Three resistors of value 3 ohm are connected in series across a potential difference of 3V

(i) Find the net current flowing through the resistor

(ii) the potential drop across a single resistor

(iii) The power dissipated in the circuit

Solution: Resistors are in series so, the total resistance = sum of the given resistors= 3+3+3= 9 ohm

Total current = V/R from Ohm's law

I=V/R=3/9=0.33A

ii) potential drop across each resistor = current through the resistor multiplied by the resistance value

$V_1=IR_1$

since the same current flows through the series resistors potential drop across $3\Omega=0.33\times3=1V$

iii) Power dissipated = $I^2R=0.33^2\times9=1\ Joule$

Q2)

Three resistors of value 3 ohm are connected in parallel across a potential difference of 3V

(i) Find the net current flowing through the resistor

(ii) the current through a single resistor

(iii) The power dissipated in the circuit

i) Resistors are in parallel therefore

$\\\frac{1}{R}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\\ \frac{1}{R}=\frac{3}{3}\\R=1\Omega$

Total current I=V/R=3/1=3A

ii) Current through a 3 Ohm resistor =I/3=3/3=1 A (since the potential difference is same across all the resistors in a parallel circuit)

iii) Power dissipated = $I^2R=3^2\times1=9\ Joule$

Certain points to remember from NCERT Solutions for Class 10 Science Chapter 12

• 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 wise

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

## NCERT Solutions for Class 10 - Subject Wise

 NCERT Solutions for Class 10 Maths NCERT Solutions for Class 10 Science