NCERT CLASS 12 PHYSICS

CHAPTER 3 – CURRENT ELECTRICITY

PART A – IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS

1. What is electric current?

Electric current is the rate of flow of electric charge through a cross-section of a conductor.

For a steady current,

I = Q/t

Where:
I = Electric current
Q = Total charge flowing through the conductor
t = Time taken

The SI unit of electric current is ampere (A).

2. Define one ampere of electric current.

One ampere is the electric current flowing through a conductor when one coulomb of charge passes through its cross-section in one second.

1 A = 1 C/s

3. What is the difference between steady current and varying current?

Steady Current: A current whose magnitude and direction remain constant with time is called steady current.

Varying Current: A current whose magnitude or direction changes with time is called varying current.

4. Why is electric current considered a scalar quantity?

Electric current is considered a scalar quantity because it is added algebraically and does not follow the laws of vector addition.

Although current is represented by an arrow to indicate its direction, it is not a vector quantity.

5. What is conventional current?

Conventional current is the direction in which positive charges would flow in a conductor.

In a metallic conductor, conventional current flows opposite to the direction of electron drift.

6. What is current density?

Current density is defined as the electric current flowing per unit area of cross-section of a conductor held perpendicular to the direction of current.

J = I/A

Where:
J = Current density
I = Electric current
A = Area of cross-section

Its SI unit is A/m².

Current density is a vector quantity.

7. What are free electrons?

Free electrons are the electrons that are not tightly bound to individual atoms and can move through a metallic conductor.

These electrons are responsible for the conduction of electric current in metals.

8. Why does a conductor not carry current in the absence of an electric field?

In the absence of an external electric field, free electrons move randomly inside a conductor due to thermal motion.

Their average velocity is zero because their motion is random in all directions.

Therefore, there is no net flow of charge and no electric current.

9. What is drift velocity?

Drift velocity is the average velocity acquired by free electrons in a conductor due to an applied electric field.

It is directed opposite to the direction of the electric field.

The magnitude of drift velocity is generally very small.

10. Why is the drift velocity of electrons very small?

The drift velocity is small because free electrons undergo frequent collisions with the positive ions of the conductor.

These collisions prevent electrons from continuously accelerating and result in a small average velocity.

11. What is the relation between electric current and drift velocity?

The electric current flowing through a conductor is given by:

I = neAvd

Where:
n = Number density of free electrons
e = Magnitude of charge of an electron
A = Area of cross-section
vd = Drift speed of electrons

12. What is relaxation time?

Relaxation time is the average time interval between two successive collisions of a conduction electron with the lattice ions of a conductor.

It is represented by τ (tau).

Its SI unit is second (s).

13. What is mobility of charge carriers?

Mobility is defined as the magnitude of drift velocity acquired by a charge carrier per unit electric field applied to the conductor.

μ = vd/E

Its SI unit is m² V⁻¹ s⁻¹.

14. State Ohm's law.

Ohm's law states that the electric current flowing through a conductor is directly proportional to the potential difference across its ends, provided its temperature and other physical conditions remain constant.

V ∝ I

V = IR

Where:
V = Potential difference
I = Electric current
R = Resistance

15. What is electrical resistance?

Electrical resistance is the property of a conductor by which it opposes the flow of electric current through it.

R = V/I

Its SI unit is ohm (Ω).

16. Define one ohm.

A conductor is said to have a resistance of one ohm if a potential difference of one volt across its ends produces a current of one ampere through it.

1 Ω = 1 V/A

17. What factors affect the resistance of a conductor?

The resistance of a conductor depends on:

Length of the conductor.
Area of cross-section.
Nature of the material.
Temperature of the conductor.

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

R = ρl/A

18. What is resistivity?

Resistivity is defined as the resistance of a conductor having unit length and unit area of cross-section.

ρ = RA/l

Its SI unit is ohm metre (Ω m).

Resistivity depends on the nature and temperature of the material, not on the dimensions of the conductor.

19. What is electrical conductivity?

Electrical conductivity is the reciprocal of electrical resistivity.

σ = 1/ρ

Its SI unit is siemens per metre (S/m).

A material having high conductivity offers less opposition to the flow of electric current.

20. What is the difference between resistance and resistivity?

Resistance:

Resistance is the opposition offered by a particular conductor to the flow of current.
It depends on the length and area of cross-section of the conductor.
Its SI unit is ohm (Ω).

Resistivity:

Resistivity is a property of the material of a conductor.
It depends on the nature and temperature of the material.
Its SI unit is ohm metre (Ω m).

21. What are ohmic conductors?

Ohmic conductors are those conductors that obey Ohm's law.

For these conductors, the potential difference is directly proportional to the current at constant temperature.

Examples: Copper wire and other metallic conductors under suitable conditions.

22. What are non-ohmic conductors?

Non-ohmic conductors are those conductors that do not obey Ohm's law.

Their current is not directly proportional to the potential difference.

Examples: Semiconductor diodes and filament lamps under changing temperature conditions.

23. What is the V-I characteristic of an ohmic conductor?

The V-I characteristic of an ohmic conductor is a straight line passing through the origin.

The slope of the V-I graph, when voltage is plotted on the vertical axis and current on the horizontal axis, gives resistance.

R = V/I

24. Why does the resistance of a metallic conductor increase with temperature?

When the temperature of a metallic conductor increases, the lattice ions vibrate more strongly.

This increases the frequency of collisions between free electrons and lattice ions.

As a result, the resistance of the metallic conductor increases.

25. How does the resistance of a semiconductor change with temperature?

The resistance of a semiconductor generally decreases with an increase in temperature.

This happens because more charge carriers become available for conduction at higher temperatures.

26. What is the temperature coefficient of resistance?

The temperature coefficient of resistance is defined as the fractional change in resistance per unit change in temperature.

α = (R₂ − R₁)/[R₁(T₂ − T₁)]

Its SI unit is K⁻¹.

27. What is electrical power?

Electrical power is the rate at which electrical energy is consumed or converted into another form of energy.

P = VI

Its SI unit is watt (W).

28. State Joule's law of heating.

Joule's law of heating states that the heat produced in a conductor due to the flow of electric current is directly proportional to:

The square of the current.
The resistance of the conductor.
The time for which current flows.

H = I²Rt

29. What is electromotive force (EMF)?

Electromotive force is the energy supplied by a cell or battery per unit charge passing through it.

ε = W/Q

Its SI unit is volt (V).

EMF is not actually a force. It represents energy supplied per unit charge.

30. What is internal resistance of a cell?

Internal resistance is the resistance offered to the flow of current inside a cell due to its electrolyte and electrodes.

It is represented by r.

Its SI unit is ohm (Ω).

31. What is terminal potential difference?

Terminal potential difference is the potential difference between the two terminals of a cell when it is connected to an external circuit.

When a cell supplies current:

V = ε − Ir

Where:
ε = EMF
I = Current
r = Internal resistance

32. What is the difference between EMF and terminal potential difference?

EMF:

It is the energy supplied by a cell per unit charge.
It is measured when no current is drawn from the cell.
It is represented by ε.

Terminal Potential Difference:

It is the potential difference between the terminals of a cell.
It is measured when the cell is connected to a circuit.
It is represented by V.

33. What happens to the terminal voltage of a cell when it supplies current?

When a cell supplies current, some potential difference is lost inside the cell due to its internal resistance.

Therefore, terminal voltage becomes less than the EMF.

V = ε − Ir

34. What happens to the terminal voltage of a cell when it is being charged?

When a cell is being charged, current flows into its positive terminal.

The terminal voltage becomes greater than the EMF.

V = ε + Ir

35. What is a combination of cells in series?

When the positive terminal of one cell is connected to the negative terminal of the next cell, the cells are said to be connected in series.

For n identical cells:

Equivalent EMF = nε

Equivalent internal resistance = nr

36. What is a combination of cells in parallel?

When all the positive terminals of cells are connected together and all the negative terminals are connected together, the cells are said to be connected in parallel.

For n identical cells:

Equivalent EMF = ε

Equivalent internal resistance = r/n

37. State Kirchhoff's junction rule.

Kirchhoff's junction rule states that the sum of currents entering a junction is equal to the sum of currents leaving the junction.

ΣI entering = ΣI leaving

This rule is based on the law of conservation of electric charge.

38. State Kirchhoff's loop rule.

Kirchhoff's loop rule states that the algebraic sum of potential differences and EMFs around any closed loop of an electrical circuit is zero.

ΣΔV = 0

This rule is based on the law of conservation of energy.

39. What is a Wheatstone bridge?

A Wheatstone bridge is an electrical circuit consisting of four resistances arranged in the form of a quadrilateral.

It is used to determine an unknown resistance accurately.

40. What is the balance condition of a Wheatstone bridge?

A Wheatstone bridge is balanced when no current flows through the galvanometer.

The balance condition is:

P/Q = R/S

Where P, Q, R and S are the four resistances of the bridge.

41. Why is a Wheatstone bridge called a null-deflection method?

A Wheatstone bridge is called a null-deflection method because, at balance, the galvanometer shows zero deflection.

This indicates that no current flows through the galvanometer.

42. Why is high voltage used for the transmission of electrical power?

Electrical power is transmitted at high voltage to reduce power loss in transmission wires.

Power loss is given by:

Ploss = I²R

For a fixed transmitted power, increasing the voltage decreases the current.

Therefore, power loss in transmission wires is reduced.

PART B – NCERT EXERCISE QUESTIONS AND ANSWERS

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?

Solution:

Given:

EMF, ε = 12 V

Internal resistance, r = 0.4 Ω

The maximum current is drawn when the external resistance is zero.

Formula:

Imax = ε/r

Substituting the values:

Imax = 12/0.4

Imax = 30 A

Answer: The maximum current that can be drawn from the 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?

Solution:

Given:

EMF, ε = 10 V

Internal resistance, r = 3 Ω

Current, I = 0.5 A

Let the external resistance be R.

Using the formula:

I = ε/(R + r)

Substituting the values:

0.5 = 10/(R + 3)

0.5(R + 3) = 10

R + 3 = 10/0.5

R + 3 = 20

R = 20 − 3

R = 17 Ω

Now, terminal voltage is:

V = IR

V = 0.5 × 17

V = 8.5 V

Answer:

Resistance of the resistor = 17 Ω

Terminal voltage of the battery = 8.5 V

QUESTION 3.3

At room temperature (27.0°C), the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10⁻⁴ °C⁻¹?

Solution:

Given:

Initial temperature, T₁ = 27°C

Initial resistance, R₁ = 100 Ω

Final resistance, R₂ = 117 Ω

Temperature coefficient, α = 1.70 × 10⁻⁴ °C⁻¹

Using the formula:

R₂ = R₁[1 + α(T₂ − T₁)]

Substituting the values:

117 = 100[1 + 1.70 × 10⁻⁴(T₂ − 27)]

117/100 = 1 + 1.70 × 10⁻⁴(T₂ − 27)

1.17 − 1 = 1.70 × 10⁻⁴(T₂ − 27)

0.17 = 1.70 × 10⁻⁴(T₂ − 27)

T₂ − 27 = 0.17/(1.70 × 10⁻⁴)

T₂ − 27 = 1000

T₂ = 1000 + 27

T₂ = 1027°C

Answer: The temperature of the heating element is 1027°C.

QUESTION 3.4

A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10⁻⁷ m², and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

Solution:

Given:

Length, l = 15 m

Area of cross-section, A = 6.0 × 10⁻⁷ m²

Resistance, R = 5.0 Ω

We know:

R = ρl/A

Therefore,

ρ = RA/l

Substituting the values:

ρ = (5.0 × 6.0 × 10⁻⁷)/15

ρ = 30 × 10⁻⁷/15

ρ = 2.0 × 10⁻⁷ Ω m

Answer: The resistivity of the material is 2.0 × 10⁻⁷ Ω m.

QUESTION 3.5

A silver wire has a resistance of 2.1 Ω at 27.5°C, and a resistance of 2.7 Ω at 100°C. Determine the temperature coefficient of resistivity of silver.

Solution:

Given:

R₁ = 2.1 Ω

T₁ = 27.5°C

R₂ = 2.7 Ω

T₂ = 100°C

We know:

R₂ = R₁[1 + α(T₂ − T₁)]

Therefore,

α = (R₂ − R₁)/[R₁(T₂ − T₁)]

Substituting the values:

α = (2.7 − 2.1)/[2.1(100 − 27.5)]

α = 0.6/(2.1 × 72.5)

α = 0.6/152.25

α = 0.00394 °C⁻¹

Approximately,

α = 3.9 × 10⁻³ °C⁻¹

Answer: The temperature coefficient of resistivity of silver is 3.9 × 10⁻³ °C⁻¹.

QUESTION 3.6

A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0°C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10⁻⁴ °C⁻¹.

Solution:

Given:

Supply voltage, V = 230 V

Initial current, I₁ = 3.2 A

Final current, I₂ = 2.8 A

Initial temperature, T₁ = 27°C

Temperature coefficient, α = 1.70 × 10⁻⁴ °C⁻¹

First, calculate the initial resistance.

R₁ = V/I₁

R₁ = 230/3.2

R₁ = 71.875 Ω

Now, calculate the final resistance.

R₂ = V/I₂

R₂ = 230/2.8

R₂ = 82.143 Ω

We know:

R₂ = R₁[1 + α(T₂ − T₁)]

Therefore,

T₂ − T₁ = (R₂ − R₁)/(αR₁)

Substituting the values:

T₂ − 27 = (82.143 − 71.875)/(1.70 × 10⁻⁴ × 71.875)

T₂ − 27 = 10.268/0.01221875

T₂ − 27 ≈ 840.3°C

T₂ = 840.3 + 27

T₂ ≈ 867.3°C

Answer: The steady temperature of the heating element is approximately 867°C.

QUESTION 3.7

Determine the current in each branch of the network shown in Fig. 3.20. The Wheatstone bridge network has AB = 10 Ω, BC = 5 Ω, CD = 10 Ω, AD = 5 Ω and BD = 15 Ω, with a 10 V battery connected across AC.

Solution:

Given:

AB = 10 Ω

BC = 5 Ω

CD = 10 Ω

AD = 5 Ω

BD = 15 Ω

Battery voltage = 10 V

Let:

Current through AB = I₁

Current through AD = I₂

Current through BD = Ig

Applying Kirchhoff's loop rule to loop ABDA:

10I₁ + 15Ig − 5I₂ = 0

Dividing by 5:

2I₁ + 3Ig − I₂ = 0 .......... (1)

Applying Kirchhoff's loop rule to loop BCDB:

5(I₁ − Ig) − 10(I₂ + Ig) − 15Ig = 0

5I₁ − 5Ig − 10I₂ − 10Ig − 15Ig = 0

5I₁ − 10I₂ − 30Ig = 0

Dividing by 5:

I₁ − 2I₂ − 6Ig = 0 .......... (2)

Applying Kirchhoff's loop rule to loop ADCA:

5I₂ + 10(I₂ + Ig) = 10

5I₂ + 10I₂ + 10Ig = 10

15I₂ + 10Ig = 10 .......... (3)

Solving equations (1), (2) and (3), we get:

I₁ = 4/17 A

I₂ = 6/17 A

Ig = −2/17 A

The negative sign indicates that the actual current through BD flows opposite to the assumed direction.

Therefore, the current through BD is:

Ig = 2/17 A, from D to B.

Current through AB:

IAB = 4/17 A

Current through BC:

IBC = I₁ − Ig

IBC = 4/17 − (−2/17)

IBC = 6/17 A

Current through AD:

IAD = 6/17 A

Current through DC:

IDC = I₂ + Ig

IDC = 6/17 − 2/17

IDC = 4/17 A

Total current supplied by the battery:

I = I₁ + I₂

I = 4/17 + 6/17

I = 10/17 A

Answer:

Current through AB = 4/17 A

Current through BC = 6/17 A

Current through AD = 6/17 A

Current through DC = 4/17 A

Current through BD = 2/17 A, from D to B

Total current = 10/17 A

QUESTION 3.8

A storage battery of EMF 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V DC supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Solution:

Given:

EMF of battery, ε = 8.0 V

Internal resistance, r = 0.5 Ω

Supply voltage = 120 V

Series resistance, R = 15.5 Ω

Net voltage driving the charging current:

Vnet = 120 − 8

Vnet = 112 V

Total resistance:

Rtotal = R + r

Rtotal = 15.5 + 0.5

Rtotal = 16 Ω

Charging current:

I = Vnet/Rtotal

I = 112/16

I = 7 A

During charging, terminal voltage is:

V = ε + Ir

V = 8 + (7 × 0.5)

V = 8 + 3.5

V = 11.5 V

Answer:

Terminal voltage of the battery during charging = 11.5 V.

Purpose of the series resistor:

The series resistor limits the charging current to a safe value and protects the battery from excessive current.

QUESTION 3.9

The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10²⁸ m⁻³. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10⁻⁶ m² and it is carrying a current of 3.0 A.

Solution:

Given:

Number density, n = 8.5 × 10²⁸ m⁻³

Length of wire, l = 3.0 m

Area of cross-section, A = 2.0 × 10⁻⁶ m²

Current, I = 3.0 A

Charge of electron, e = 1.6 × 10⁻¹⁹ C

We know:

I = neAvd

Therefore,

vd = I/(neA)

Substituting the values:

vd = 3.0/[(8.5 × 10²⁸)(1.6 × 10⁻¹⁹)(2.0 × 10⁻⁶)]

vd = 3.0/(2.72 × 10⁴)

vd = 1.103 × 10⁻⁴ m/s

Now,

Time = Distance/Speed

t = l/vd

t = 3.0/(1.103 × 10⁻⁴)

t ≈ 2.72 × 10⁴ s

Converting into hours:

t = 27200/3600

t ≈ 7.55 hours

Answer: The time taken by an electron to drift from one end of the wire to the other is approximately 2.7 × 10⁴ seconds or 7.5 hours.

PART C – IMPORTANT QUESTIONS AND ANSWERS

1. Derive the relation between electric current and drift velocity.

Answer:

Consider a conductor having:

A = Area of cross-section

n = Number of free electrons per unit volume

e = Charge of an electron

vd = Drift velocity

Let the electrons move through the conductor with drift velocity vd.

In time t, the distance travelled by electrons is:

l = vd × t

The volume of the conductor through which electrons move is:

Volume = A × vd × t

Number of free electrons in this volume:

N = nAvdt

Total charge flowing through the cross-section:

Q = Ne

Q = nAvdte

We know:

I = Q/t

Therefore,

I = nAvdte/t

I = neAvd

Hence proved.

2. Derive the expression for drift velocity of electrons.

Answer:

Consider a free electron of charge −e and mass m moving inside a conductor.

When an electric field E is applied, the force acting on the electron is:

F = −eE

According to Newton's second law:

F = ma

−eE = ma

Therefore,

a = −eE/m

If τ is the average time between two successive collisions, the average drift velocity is:

vd = aτ

Substituting the value of acceleration:

vd = (−eE/m)τ

vd = −eEτ/m

The negative sign indicates that electrons drift opposite to the direction of the electric field.

Magnitude:

vd = eEτ/m

Hence proved.

3. Derive the expression for resistance of a conductor.

Answer:

Consider a conductor of:

Length = l

Area of cross-section = A

Resistance = R

Experimentally, resistance is directly proportional to the length of the conductor.

R ∝ l .......... (1)

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

R ∝ 1/A .......... (2)

Combining equations (1) and (2):

R ∝ l/A

Removing the proportionality sign:

R = ρl/A

Where ρ is the resistivity of the material.

Hence proved.

4. Derive the relation between resistivity and relaxation time.

Answer:

We know that drift velocity is:

vd = eEτ/m

Current density is:

J = nevd

Substituting the value of drift velocity:

J = ne(eEτ/m)

J = ne²Eτ/m

According to microscopic Ohm's law:

J = σE

Comparing the two expressions:

σE = ne²Eτ/m

Therefore,

σ = ne²τ/m

We know:

ρ = 1/σ

Hence,

ρ = m/(ne²τ)

This is the required relation between resistivity and relaxation time.

5. Derive the relation between electrical conductivity and electric field.

Answer:

We know:

vd = eEτ/m

Current density:

J = nevd

Substituting drift velocity:

J = ne(eEτ/m)

J = (ne²τ/m)E

Comparing with:

J = σE

We get:

σ = ne²τ/m

Therefore,

J = σE

This is the microscopic form of Ohm's law.

6. Explain the variation of resistance with temperature.

Answer:

The resistance of a conductor changes with temperature.

For a metallic conductor, resistance generally increases with an increase in temperature.

The relation is:

R₂ = R₁[1 + α(T₂ − T₁)]

Where:

R₁ = Resistance at initial temperature T₁

R₂ = Resistance at final temperature T₂

α = Temperature coefficient of resistance

For metals, α is positive.

For semiconductors, resistance generally decreases with an increase in temperature.

For many alloys, the temperature coefficient is comparatively small.

7. Derive the expression for electrical power.

Answer:

Electrical power is the rate of consumption of electrical energy.

P = W/t

We know that electrical work done is:

W = VQ

Therefore,

P = VQ/t

But,

I = Q/t

Hence,

P = VI

Using Ohm's law:

V = IR

Therefore,

P = I²R

Also,

I = V/R

Therefore,

P = V²/R

Thus, the three expressions for electrical power are:

P = VI = I²R = V²/R

8. Explain Joule's law of heating.

Answer:

According to Joule's law of heating, the heat produced in a conductor is directly proportional to the square of the current, the resistance of the conductor and the time for which current flows.

Mathematically,

H = I²Rt

Where:

H = Heat produced

I = Current

R = Resistance

t = Time

The SI unit of heat energy is joule (J).

9. Explain EMF and internal resistance of a cell.

Answer:

EMF:

EMF is the energy supplied by a cell per unit charge.

ε = W/Q

Its SI unit is volt.

Internal Resistance:

The resistance offered by the electrolyte and electrodes inside a cell is called internal resistance.

It is represented by r.

When a cell supplies current:

V = ε − Ir

Where V is terminal voltage.

10. Derive the relation between EMF and terminal voltage of a cell.

Answer:

Consider a cell of EMF ε and internal resistance r connected to an external resistance R.

Let I be the current flowing through the circuit.

According to Ohm's law, the potential difference across the external resistance is:

V = IR

The total resistance of the circuit is:

R + r

Therefore,

I = ε/(R + r)

Multiplying both sides by R + r:

ε = IR + Ir

Since IR = V,

ε = V + Ir

Therefore,

V = ε − Ir

This is the relation between terminal voltage and EMF when a cell supplies current.

11. Derive the equivalent EMF and internal resistance of cells connected in series.

Answer:

Consider n identical cells connected in series.

Let each cell have:

EMF = ε

Internal resistance = r

In a series combination, the total EMF is the sum of the individual EMFs.

Equivalent EMF:

εeq = ε + ε + ε + ... n times

εeq = nε

The total internal resistance is also the sum of individual internal resistances.

req = nr

Therefore, for n identical cells connected in series:

Equivalent EMF = nε

Equivalent internal resistance = nr

12. Derive the equivalent EMF and internal resistance of identical cells connected in parallel.

Answer:

Consider n identical cells connected in parallel.

Let each cell have:

EMF = ε

Internal resistance = r

In parallel combination, the equivalent EMF remains equal to the EMF of one cell.

Therefore,

εeq = ε

The equivalent internal resistance is:

1/req = 1/r + 1/r + ... n times

1/req = n/r

Therefore,

req = r/n

Hence, equivalent EMF is ε and equivalent internal resistance is r/n.

13. State and explain Kirchhoff's laws.

Answer:

Kirchhoff's laws are used to analyse complex electrical circuits.

First Law – Junction Rule:

The sum of currents entering a junction is equal to the sum of currents leaving the junction.

ΣI entering = ΣI leaving

It is based on conservation of electric charge.

Second Law – Loop Rule:

The algebraic sum of potential differences and EMFs in a closed loop is zero.

ΣΔV = 0

It is based on conservation of energy.

14. Explain the working principle of a Wheatstone bridge.

Answer:

A Wheatstone bridge consists of four resistances connected in the form of a quadrilateral.

A galvanometer is connected between the two middle junctions, and a cell is connected across the opposite junctions.

When the bridge is balanced, no current flows through the galvanometer.

The balance condition is:

P/Q = R/S

This condition is used to determine an unknown resistance.

15. Why is electrical power transmitted at high voltage?

Answer:

During transmission, electrical energy is lost in the form of heat due to the resistance of transmission wires.

Power loss is:

Ploss = I²R

For a fixed amount of power:

P = VI

Therefore,

I = P/V

Substituting:

Ploss = (P/V)²R

Ploss = P²R/V²

Thus,

Ploss ∝ 1/V²

Therefore, transmitting electrical power at high voltage reduces current and minimises power loss.

PART D – IMPORTANT NUMERICAL QUESTIONS

1. A current of 2 A flows through a conductor for 5 minutes. Calculate the total charge passing through the conductor.

Solution:

Given:

I = 2 A

t = 5 × 60 = 300 s

We know:

Q = It

Q = 2 × 300

Q = 600 C

Answer: 600 C

2. A wire has a resistance of 10 Ω. If a potential difference of 20 V is applied across it, calculate the current flowing through the wire.

Solution:

Given:

R = 10 Ω

V = 20 V

Using Ohm's law:

I = V/R

I = 20/10

I = 2 A

Answer: 2 A

3. A wire of length 2 m and area of cross-section 1 × 10⁻⁶ m² has a resistivity of 2 × 10⁻⁸ Ω m. Calculate its resistance.

Solution:

Given:

l = 2 m

A = 1 × 10⁻⁶ m²

ρ = 2 × 10⁻⁸ Ω m

Using:

R = ρl/A

R = (2 × 10⁻⁸ × 2)/(1 × 10⁻⁶)

R = 4 × 10⁻² Ω

Answer: 0.04 Ω

4. A conductor of resistance 5 Ω carries a current of 3 A for 10 seconds. Calculate the heat produced.

Solution:

Given:

R = 5 Ω

I = 3 A

t = 10 s

Using Joule's law:

H = I²Rt

H = 3² × 5 × 10

H = 9 × 50

H = 450 J

Answer: 450 J

5. A cell of EMF 12 V and internal resistance 1 Ω is connected to an external resistance of 5 Ω. Calculate the current and terminal voltage.

Solution:

Given:

ε = 12 V

r = 1 Ω

R = 5 Ω

Current:

I = ε/(R + r)

I = 12/(5 + 1)

I = 12/6

I = 2 A

Terminal voltage:

V = ε − Ir

V = 12 − (2 × 1)

V = 10 V

Answer: Current = 2 A; Terminal voltage = 10 V.

6. Three identical cells, each of EMF 2 V and internal resistance 0.5 Ω, are connected in series. Calculate the equivalent EMF and equivalent internal resistance.

Solution:

Given:

n = 3

ε = 2 V

r = 0.5 Ω

Equivalent EMF:

εeq = nε

εeq = 3 × 2

εeq = 6 V

Equivalent internal resistance:

req = nr

req = 3 × 0.5

req = 1.5 Ω

Answer: Equivalent EMF = 6 V; Equivalent internal resistance = 1.5 Ω.

7. Three identical cells, each having an internal resistance of 6 Ω, are connected in parallel. Calculate their equivalent internal resistance.

Solution:

Given:

n = 3

r = 6 Ω

For identical cells in parallel:

req = r/n

req = 6/3

req = 2 Ω

Answer: 2 Ω

PART E – IMPORTANT FORMULAS

Electric current:

I = Q/t

Charge:

Q = It

Current density:

J = I/A

Ohm's law:

V = IR

Resistance:

R = V/I

Resistance of a conductor:

R = ρl/A

Resistivity:

ρ = RA/l

Conductivity:

σ = 1/ρ

Relation between current and drift velocity:

I = neAvd

Drift velocity:

vd = eEτ/m

Mobility:

μ = vd/E

Resistivity in terms of relaxation time:

ρ = m/(ne²τ)

Temperature dependence of resistance:

R₂ = R₁[1 + α(T₂ − T₁)]

Electrical power:

P = VI

Electrical power:

P = I²R

Electrical power:

P = V²/R

Joule's law of heating:

H = I²Rt

EMF:

ε = W/Q

Current through a cell:

I = ε/(R + r)

Terminal voltage during discharge:

V = ε − Ir

Terminal voltage during charging:

V = ε + Ir

Maximum current from a cell:

Imax = ε/r

Cells in series:

εeq = nε

req = nr

Identical cells in parallel:

εeq = ε

req = r/n

Kirchhoff's junction rule:

ΣI entering = ΣI leaving

Kirchhoff's loop rule:

ΣΔV = 0

Wheatstone bridge balance condition:

P/Q = R/S

Power loss in transmission wires:

Ploss = I²R

PART F – IMPORTANT SI UNITS

Physical Quantity	SI Unit
Electric current	Ampere (A)
Electric charge	Coulomb (C)
Potential difference	Volt (V)
EMF	Volt (V)
Resistance	Ohm (Ω)
Resistivity	Ohm metre (Ω m)
Conductivity	Siemens per metre (S/m)
Current density	A/m²
Drift velocity	m/s
Relaxation time	Second (s)
Mobility	m² V⁻¹ s⁻¹
Electrical power	Watt (W)
Electrical energy	Joule (J)
Internal resistance	Ohm (Ω)

PART G – QUICK REVISION POINTS

Electric current is the rate of flow of electric charge.
The SI unit of current is ampere.
Conventional current flows opposite to electron drift in metals.
Current is a scalar quantity.
Current density is a vector quantity.
Drift velocity is the average velocity acquired by free electrons due to an electric field.
Ohm's law is V = IR.
Resistance is directly proportional to length and inversely proportional to area.
Resistivity is a property of the material.
Conductivity is the reciprocal of resistivity.
Mobility is drift velocity per unit electric field.
For metals, resistance generally increases with temperature.
For semiconductors, resistance generally decreases with temperature.
Electrical power is P = VI.
Joule's law of heating is H = I²Rt.
EMF is energy supplied per unit charge.
Internal resistance is the resistance inside a cell.
Terminal voltage during discharge is V = ε − Ir.
Terminal voltage during charging is V = ε + Ir.
Kirchhoff's junction rule is based on conservation of charge.
Kirchhoff's loop rule is based on conservation of energy.
A balanced Wheatstone bridge has zero current through its galvanometer.
The Wheatstone bridge balance condition is P/Q = R/S.
High-voltage transmission reduces power loss in transmission wires.