Class 12 Physics Notes – Chapter 3: Current Electricity

1. Electric Current

The rate of flow of electric charge through any cross-section of a conductor is called electric current.

If a net charge Q flows across any cross-section in time t, then current I is given by:

Formula: I = Q / t

For a varying current, instantaneous current is: I = ΔQ / Δt (as Δt → 0).

SI Unit: Ampere (A). Current is a scalar quantity.

2. Ohm’s Law

Ohm’s law states that the physical conditions (like temperature, mechanical strain, etc.) remaining unchanged, the current flowing through a conductor is directly proportional to the potential difference across its ends.

Formula: V ∝ I ⇒ V = I · R

Where R is the resistance of the conductor. Its SI unit is Ohm (Ω).

Resistance and Resistivity

The resistance R of a conductor depends on its length (l) and cross-sectional area (A):

R = ρ · (l / A)

Here, ρ (rho) is the specific resistance or resistivity of the material. Its SI unit is Ω m.

3. Drift of Electrons and Mobility

In a conductor, free electrons move randomly. When an external electric field is applied, they experience a force and attain a small average velocity opposite to the direction of the field. This is called drift velocity (v_d).

Formula: v_d = (e · E · τ) / m

Where e = charge of electron, E = electric field, τ = relaxation time, m = mass of electron.

Relation between Current and Drift Velocity:
I = n · e · A · v_d
(where n is the number density of free electrons)

Mobility (μ)

Mobility is defined as the magnitude of drift velocity per unit electric field: μ = v_d / E

4. Limitations of Ohm’s Law

Materials that do not obey Ohm’s law are called non-ohmic devices. Their V-I graph is not a straight line.

• V ceases to be proportional to I (e.g., semiconductor diodes).
• The relation between V and I depends on the sign of V.
• The relation between V and I is not unique (e.g., Gallium Arsenide).

5. Temperature Dependence of Resistivity

The resistivity of a material depends on temperature. For a limited range of temperatures, it is given by:

Formula: ρ_T = ρ₀ [1 + α(T – T₀)]

Where ρ_T and ρ₀ are resistivities at temperatures T and T₀ respectively. α is the temperature coefficient of resistivity.

• For metals: α is positive (resistivity increases with temperature).
• For semiconductors and insulators: α is negative (resistivity decreases with temperature).

6. Electrical Energy and Power

The rate at which electrical energy is dissipated into other forms of energy (like heat) is called electrical power (P).

Formulas: P = V · I = I²R = V² / R

The SI unit of power is Watt (W).

7. Cells, EMF, and Internal Resistance

Electromotive Force (emf, ε): The potential difference between the two poles of a cell in an open circuit (when no current is drawn).

Internal Resistance (r): The resistance offered by the electrolyte and electrodes of a cell.

Terminal Voltage (V): When a current I is drawn from a cell, the voltage across its terminals drops: V = ε – I · r

Current drawn from a cell connected to external resistance R: I = ε / (R + r)

8. Kirchhoff’s Rules

• Junction Rule (KCL): At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction. (Based on conservation of charge). ΣI = 0.
• Loop Rule (KVL): The algebraic sum of changes in potential around any closed loop involving resistors and cells in the circuit is zero. (Based on conservation of energy). ΣΔV = 0.

9. Wheatstone Bridge

It is an arrangement of four resistors R₁, R₂, R₃, and R₄ in the form of a bridge. When the galvanometer shows no deflection, the bridge is balanced.

Balanced Condition: R₂ / R₁ = R₄ / R₃

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Multiple Choice Questions (MCQs)

1. The SI unit of electric current is:
a) Coulomb
b) Volt
c) Ampere
d) Ohm
Answer: (c) Ampere

2. According to Ohm’s Law, the relationship between Voltage (V) and Current (I) is:
a) V = I / R
b) V = I · R
c) V = R / I
d) V = I²R
Answer: (b) V = I · R

3. The mobility (μ) of charge carriers is given by the formula:
a) μ = v_d · E
b) μ = E / v_d
c) μ = v_d / E
d) μ = e / m
Answer: (c) μ = v_d / E

4. For metallic conductors, what happens to their resistance when temperature increases?
a) Decreases
b) Remains constant
c) Increases
d) Becomes zero
Answer: (c) Increases

5. The SI unit of resistivity (ρ) is:
a) Ω
b) Ω m
c) Ω^-1 m^-1
d) Ω / m
Answer: (b) Ω m

6. Kirchhoff’s Junction Rule is a consequence of the law of conservation of:
a) Energy
b) Momentum
c) Mass
d) Charge
Answer: (d) Charge

7. Kirchhoff’s Loop Rule is based on the law of conservation of:
a) Energy
b) Charge
c) Mass
d) Momentum
Answer: (a) Energy

8. The condition for a balanced Wheatstone bridge with arms R₁, R₂, R₃, R₄ is:
a) R₁ · R₂ = R₃ · R₄
b) R₁ + R₂ = R₃ + R₄
c) R₂ / R₁ = R₄ / R₃
d) R₁ – R₂ = R₃ – R₄
Answer: (c) R₂ / R₁ = R₄ / R₃

9. The electrical power dissipated in a resistor is given by:
a) V · R
b) I²R
c) I / R
d) I² / R
Answer: (b) I²R

10. The relation between terminal voltage (V), emf (ε), current (I), and internal resistance (r) during discharge is:
a) V = ε + I · r
b) V = ε – I · r
c) V = I · r – ε
d) V = ε / (I · r)
Answer: (b) V = ε – I · r

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Textbook Exercise Solutions

Q 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.
Maximum current is drawn when external resistance R = 0.
I_max = ε / r = 12 / 0.4 = 30 A.

Q 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: ε = 10 V, r = 3 Ω, I = 0.5 A.
(a) Current I = ε / (R + r)
0.5 = 10 / (R + 3) ⇒ R + 3 = 10 / 0.5 = 20
R = 20 – 3 = 17 Ω.
(b) Terminal voltage V = I · R = 0.5 × 17 = 8.5 V.
(Alternatively, V = ε – I · r = 10 – (0.5 × 3) = 10 – 1.5 = 8.5 V).

Q 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: T₁ = 27.0 °C, R₁ = 100 Ω, R₂ = 117 Ω, α = 1.70 × 10^-4 °C^-1.
Using formula: R₂ = R₁ [1 + α(T₂ – T₁)]
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) = 1000
T₂ = 1000 + 27 = 1027 °C.

Q 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: l = 15 m, A = 6.0 × 10^-7 m², R = 5.0 Ω.
Resistivity ρ = (R · A) / l
ρ = (5.0 × 6.0 × 10^-7) / 15
ρ = 30.0 × 10^-7 / 15 = 2.0 × 10^-7 Ω m.

Q 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: T₁ = 27.5 °C, R₁ = 2.1 Ω; T₂ = 100 °C, R₂ = 2.7 Ω.
Formula: α = (R₂ – R₁) / [R₁(T₂ – T₁)]
α = (2.7 – 2.1) / [2.1 × (100 – 27.5)]
α = 0.6 / [2.1 × 72.5] = 0.6 / 152.25
α ≈ 0.0039 °C^-1.

Q 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: V = 230 V, I₁ = 3.2 A, I₂ = 2.8 A, T₁ = 27.0 °C, α = 1.70 × 10^-4 °C^-1.
Initial Resistance R₁ = V / I₁ = 230 / 3.2 = 71.875 Ω.
Steady Resistance R₂ = V / I₂ = 230 / 2.8 = 82.143 Ω.
Using α = (R₂ – R₁) / [R₁(T₂ – T₁)]
T₂ – T₁ = (R₂ – R₁) / (R₁ · α)
T₂ – 27 = (82.143 – 71.875) / (71.875 × 1.70 × 10^-4)
T₂ – 27 = 10.268 / 0.01221875 ≈ 840.3
T₂ = 840.3 + 27 = 867.3 °C.

Q 3.7: Determine the current in each branch of the network shown in Fig. 3.20 (A Wheatstone bridge with AB=10 Ω, BC=5 Ω, CD=5 Ω, DA=5 Ω, BD=5 Ω and a 10V battery across AC).
Solution:
Applying Kirchhoff’s loop rules to the closed loops ABDA and BCDB, and solving the linear equations for the currents, we get:
Total current from the battery, I = 10/17 A.
Current in branch AB, I_AB = 4/17 A
Current in branch BC, I_BC = 6/17 A
Current in branch CD, I_CD = -4/17 A (or 4/17 A from C to D)
Current in branch AD, I_AD = 6/17 A
Current through Galvanometer branch BD, I_BD = -2/17 A (or 2/17 A from D to B).

Q 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: V_supply = 120 V, emf ε = 8.0 V, r = 0.5 Ω, R = 15.5 Ω.
During charging, the net emf is (V_supply – ε).
Current I = (V_supply – ε) / (R + r) = (120 – 8.0) / (15.5 + 0.5) = 112 / 16.0 = 7.0 A.
Terminal voltage during charging, V = ε + I · r
V = 8.0 + (7.0 × 0.5) = 8.0 + 3.5 = 11.5 V.
Purpose of series resistor: It limits the high current drawn from the external supply. Without it, the charging current would be dangerously high.

Q 3.9: The number density of free electrons in a copper conductor 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: n = 8.5 × 10²⁸ m^-3, l = 3.0 m, A = 2.0 × 10^-6 m², I = 3.0 A, e = 1.6 × 10^-19 C.
We know, I = n · e · A · v_d ⇒ v_d = I / (n · e · A)
Time taken t = distance / velocity = l / v_d = l / [I / (n · e · A)] = (l · n · e · A) / I
t = (3.0 × 8.5 × 10²⁸ × 1.6 × 10^-19 × 2.0 × 10^-6) / 3.0
t = 8.5 × 1.6 × 2.0 × 10³ = 27.2 × 10³ s = 2.72 × 10⁴ seconds (approx 7.5 hours).