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## NCERT Class 12 Physics Textbook Chapter 3 With Answer PDF Free Download

### Chapter 3: Current Electricity

#### 3.1 INTRODUCTION

In Chapter 1, all charges whether free or bound were considered to be at rest. Charges in motion constitute an electric current.

Such currents occur naturally in many situations. Lightning is one such phenomenon in which charges flow from the clouds to the earth through the atmosphere, sometimes with disastrous results.

The flow of charges in lightning is not steady, but in our everyday life, we see many devices where charges flow in a steady manner, like water flowing smoothly in a river.

A torch and a cell-driven clock are examples of such devices. In the present chapter, we shall study some of the basic laws concerning steady electric currents.

#### 3.2 Electric Current

Imagine a small area held normal to the direction of the flow of charges. Both the positive and the negative charges may flow forward and backward across the area.

In a given time interval t, let q+ be the net amount (i.e., forward minus backward) of positive charge that flows in the forward direction across the area.

Similarly, let q– be the net amount of negative charge flowing across the area in the forward direction.

The net amount of charge flowing across the area in the forward direction in the time interval t, then, is q = q+ – q– This is proportional to t for a steady current.

3.3 Electric Currents In Conductors

An electric charge will experience a force if an electric field is applied. If it is free to move, it will thus move contributing to a current.

In nature, free-charged particles do exist in the upper strata of the atmosphere called the ionosphere. However, in atoms and molecules, the negatively charged electrons and the positively charged nuclei are bound to each other and are thus not free to move.

The bulk matter is made up of many molecules, a gram of water, for example, contains approximately 1022 molecules.

These molecules are so closely packed that the electrons are no longer attached to individual nuclei.

In some materials, the electrons will still be bound, i.e., they will not accelerate even if an electric field is applied.

In other materials, notably metals, some of the electrons are practically free to move within the bulk material.

These materials, generally called conductors, develop electric currents in them when an electric field is applied. If we consider solid conductors, then of course the atoms are tightly bound to each other so that the current is carried by the negatively charged electrons.

There are, however, other types of conductors like electrolytic solutions were positive and negative charges both can move.

In our discussions, we will focus only on solid conductors so that the current is carried by the negatively charged electrons in the background of fixed positive ions.

Consider first the case when no electric field is present. The electrons will be moving due to thermal motion during which they collide with the fixed ions.

An electron colliding with an ion emerges at the same speed as before the collision.

However, the direction of its velocity after the collision is completely random. At a given time, there is no preferential

direction for the velocities of the electrons.

Author | NCERT |

Language | English |

No. of Pages | 39 |

PDF Size | 2.9 MB |

Category | English Book |

Source/Credits | ncert.nic.in |

### NCERT Solutions Class 12 Physics Chapter 3 Current Electricity

*Question 3.1:*

*The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?*

*Answer 3.1:*

In the given question,

The EMF of the battery is given as E = 12 V

The internal resistance of the battery is given as R = 0.4 Ω

The amount of maximum current drawn from the battery is given by = I

According to Ohm’s law,

E = IR

Rearranging, we getI = \frac{ E }{ R }*I*=*R**E*

Substituting values in the above equation, we getI = \frac{ 12 }{ 0.4 }*I*=0.412 = 30 A

Therefore, the maximum current drawn from the given battery is 30 A.

*Question 3.2:*

*A battery of EMF 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?*

** Answer 3.2**:

**Given:**

The EMF of the battery (* E* = 10 V)

The internal resistance of the battery (* R* = 3 Ω)

The current in the circuit (* I* = 0.5 A)

Consider the resistance of the resistor to be * R*.

The current in the circuit can be found out using Ohm’s Law as,I = \frac{ E }{ R + r }*I*=*R*+*r**E*

Rewriting the above equation, we getR + r = \frac{E}{I}*R*+*r*=*I**E* = \frac{10}{0.5} = 20 Ω=0.510=20Ω

Therefore,R = 20 – 3 = 17 Ω*R*=20–3=17Ω

Consider the Terminal voltage of the resistor to be V.

Then, according to Ohm’s law,

V = IR

Substituting values in the equation, we get

V = 0.5 × 17

V = 8.5 V

Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is 8.5 V.

*Question 3.3 :*

*a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?*

*b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.*

*Answer 3.3 :*

a) We know that resistors r _{1} = 1 Ω , r _{2} = 2 Ω and r _{3} = 3 Ω are combined in series.

The total resistance of the above series combination can be calculated by the algebraic sum of individual resistances as follows:

Total resistance = 1 Ω + 2 Ω + 3 Ω = 6 Ω

Thus calculated Total Resistance = 6 Ω

b) Let us consider I to be the current flowing the given circuit

Also,

The emf of the battery is E = 12 V

Total resistance of the circuit ( calculated above ) = R = 6 Ω

Using Ohm’s law, relation for current can be obtained asI = \frac{ E }{ R }*I*=*R**E*

Substituting values in the above equation, we getI = \frac{ 12 }{ 6 }*I*=612 = 2 A

Therefore, the current calculated is 2 A

Let the Potential drop across 1 Ω resistor = V _{1}

The value of V _{1} can be obtained from Ohm’s law as :

V _{1} = 2 x 1 = 2 V

Let the Potential drop across 2 Ω resistor = V _{2}

The value of V _{2} can be obtained from Ohm’s law as :

V _{2} = 2 x 2 = 4 V

Let the Potential drop across 3 Ω resistor = V _{3}

The value of V _{3} can be obtained from Ohm’s law as :

V _{3} = 2 x 3 = 6 V

Therefore, the potential drops across the given resistors r _{1} = 1 Ω , r _{2} = 2 Ω and r _{3} = 3 Ω are calculated to be

V _{1} = 2 x 1 = 2 V

V _{2} = 2 x 1 = 4 V

V _{3} = 2 x 1 = 6 V

*Question 3.4 :*

*a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?*

*b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.*

*Answer 3.4 :*

A ) Resistors r _{1} = 2 Ω , r_{2} = 4 Ω and r _{3} = 5 Ω are combined in parallel

Hence the total resistance of the above circuit can be calculated by the following formula :\frac{ 1 }{ R } = \frac{ 1 }{ R_1 } + \frac{ 1 }{ R_2 } + \frac{ 1 }{ R_3 }*R*1=*R*11+*R*21+*R*31\frac{ 1 }{ R } = \frac{ 1 }{ 2 } + \frac{ 1 }{ 4 } + \frac{ 1 }{ 5 }*R*1=21+41+51\frac{ 1 }{ R } = \frac{ 10 + 5 + 4 }{ 20 }*R*1=2010+5+4\frac{ 1 }{ R } = \frac{ 19 }{ 20 }*R*1=2019

Therefore, the total resistance of the parallel combination given above is given by:R = \frac{20}{19}*R*=1920

B ) Given that emf of the battery, E = 20 V

Let the current flowing through resistor R _{1} be I _{1}

I _{1} is given by :I_{ 1 } = \frac{ V }{ R_{ 1 }}*I*1=*R*1*V*I_{ 1 } = \frac{ 20 }{ 2 }*I*1=220I_{ 1 } = 10 A*I*1=10*A*

Let the current flowing through resistor R _{2} be I _{2}

I _{2} is given by :I_{ 2 } = \frac{ V }{ R_{ 2 }}*I*2=*R*2*V*I_{ 2 } = \frac{ 20 }{ 4 }*I*2=420I_{ 2 } = 5 A*I*2=5*A*

Let the current flowing through resistor R _{3} be I _{3}

I _{3} is given by :I_{ 3 } = \frac{ V }{ R_{ 1 }}*I*3=*R*1*V*I_{ 3 } = \frac{ 20 }{ 5 }*I*3=520I_{ 3 } = 4 A*I*3=4*A*

Therefore, the total current can be found by the following formula :

I = I _{1} + I _{2} + I _{3} = 10 + 5 + 4 = 19 A

therefore the current flowing through each resistor is calculated to be :I_{ 1 } = 10 A*I*1=10*A*I_{ 2 } = 5 A*I*2=5*A*I_{ 3 } = 4 A*I*3=4*A*

Therefore, the total current is I = 19 A

NCERT Class 12 Physics Textbook Chapter 3 With Answer PDF Free Download