Current Electricity Class 12 Notes | CBSE Physics Chapter 3

Current Electricity is Chapter 3 of CBSE Class 12 Physics. This chapter deals with the flow of electric charge through conductors — Ohm’s law, resistance, resistivity, Kirchhoff’s laws, Wheatstone bridge, and the meter bridge. It also covers the internal resistance of cells and combinations of cells.

This chapter is extremely important — expect 7–8 marks. Kirchhoff’s laws, Wheatstone bridge numericals, and internal resistance problems are the most tested topics.

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Key Concepts

1. Electric Current

Electric current is the rate of flow of charge: I = dQ/dt

Unit: Ampere (A). 1 A = 1 C/s

Conventional current flows from higher potential (+) to lower potential (−). Electron flow is in the opposite direction.

Drift Velocity

When a potential difference is applied, free electrons in a conductor drift slowly towards the positive terminal. This average velocity is called drift velocity (vd).

I = neAvd

• n = number density of free electrons (per m³)
• e = charge of electron
• A = cross-sectional area
• vd = drift velocity (typically ~10⁻⁴ m/s — very slow!)

Current Density

J = I/A = nevd (unit: A/m²)

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2. Ohm’s Law and Resistance

Ohm’s Law: V = IR (at constant temperature)

Resistance: R = V/I = ρl/A

• ρ = resistivity of the material (unit: Ω·m)
• l = length of conductor
• A = cross-sectional area

Temperature Dependence

For metals: ρ = ρ₀(1 + αΔT) — resistivity increases with temperature

For semiconductors: resistivity decreases with temperature (more carriers generated)

Colour Code for Resistors

Bands: Black(0), Brown(1), Red(2), Orange(3), Yellow(4), Green(5), Blue(6), Violet(7), Grey(8), White(9)

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3. Combinations of Resistors

Feature	Series	Parallel
Current	Same	Divides
Voltage	Divides	Same
Equivalent R	R = R₁ + R₂ + …	1/R = 1/R₁ + 1/R₂ + …

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4. Cells and Internal Resistance

A real cell has an EMF (ε) and an internal resistance (r).

Terminal voltage: V = ε − Ir (when current is drawn)

When no current flows (open circuit): V = ε

Combination of Cells

Type	EMF	Internal Resistance	Best for
Series (n cells)	nε	nr	High EMF needed (external R >> internal r)
Parallel (n cells)	ε	r/n	More current needed (external R << internal r)

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5. Kirchhoff’s Laws

Junction Rule (KCL — Kirchhoff’s Current Law)

The sum of currents entering a junction equals the sum of currents leaving it. (Based on conservation of charge)

ΣI_in = ΣI_out

Loop Rule (KVL — Kirchhoff’s Voltage Law)

The algebraic sum of potential differences around any closed loop is zero. (Based on conservation of energy)

ΣV = 0 around any closed loop

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6. Wheatstone Bridge

A Wheatstone bridge is a circuit with four resistors arranged in a diamond shape. When the bridge is balanced, no current flows through the galvanometer.

Balance condition: P/Q = R/S

7. Meter Bridge

A practical form of Wheatstone bridge using a 1 m wire. At balance:

R/S = l/(100 − l)

where l is the balancing length from one end.

8. Potentiometer

A device for measuring EMF accurately (draws no current from the source).

• Compares EMFs: ε₁/ε₂ = l₁/l₂
• Measures internal resistance: r = R(l₁ − l₂)/l₂

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Important Definitions

Term	Definition
Electric current	Rate of flow of charge: I = dQ/dt
Drift velocity	Average velocity of free electrons in a conductor under applied field
Resistivity	Material property that determines resistance: R = ρl/A
EMF	Work done per unit charge by the cell in moving charge through the complete circuit
Internal resistance	Resistance offered by the electrolyte and electrodes inside the cell
Kirchhoff’s junction rule	Sum of currents at a junction = 0 (conservation of charge)
Kirchhoff’s loop rule	Sum of potential differences in a closed loop = 0
Wheatstone bridge	Circuit of four resistors; balanced when P/Q = R/S

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Solved Examples

Example 1

A cell of EMF 2 V and internal resistance 0.5 Ω is connected to a 3.5 Ω resistor. Find the current and terminal voltage.

Answer: I = ε/(R + r) = 2/(3.5 + 0.5) = 2/4 = 0.5 A. Terminal voltage: V = ε − Ir = 2 − 0.5 × 0.5 = 1.75 V.

Example 2

In a Wheatstone bridge, P = 100 Ω, Q = 200 Ω, R = 150 Ω. Find S for balance.

Answer: P/Q = R/S → 100/200 = 150/S → S = 150 × 200/100 = 300 Ω.

Example 3

In a meter bridge, the null point is at 40 cm. If R = 10 Ω, find S.

Answer: R/S = l/(100 − l) → 10/S = 40/60 → S = 10 × 60/40 = 15 Ω.

Example 4

A copper wire of length 2 m and cross-section 1 mm² has resistivity 1.7 × 10⁻⁸ Ω·m. Find its resistance.

Answer: R = ρl/A = (1.7 × 10⁻⁸ × 2)/(1 × 10⁻⁶) = 3.4 × 10⁻⁸/10⁻⁶ = 0.034 Ω.

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Important Questions for Board Exams

1-Mark Questions

1. State Kirchhoff’s junction rule.
2. What is the condition for balance in a Wheatstone bridge?
3. Why is a potentiometer preferred over a voltmeter for measuring EMF?

3-Mark Questions

1. Derive the relation I = neAvd for drift velocity.
2. State and explain Kirchhoff’s laws. Use them to find the current in a given circuit.
3. Explain how a meter bridge works. Derive the formula for unknown resistance.

5-Mark Questions

1. Define resistivity. Derive the expression for equivalent resistance in series and parallel combinations.
2. Explain the working of a potentiometer. How can it be used to compare EMFs of two cells?

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Quick Revision Points

• I = dQ/dt; I = neAvd; drift velocity is very slow (~10⁻⁴ m/s)
• Ohm’s law: V = IR; Resistance: R = ρl/A
• Temperature: metals R↑, semiconductors R↓
• Cell: V = ε − Ir; Series: nε, nr; Parallel: ε, r/n
• KCL: ΣI = 0 at junction; KVL: ΣV = 0 in loop
• Wheatstone balance: P/Q = R/S
• Meter bridge: R/S = l/(100 − l)
• Potentiometer: ε₁/ε₂ = l₁/l₂ (no current drawn — accurate)

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Previous Chapter: Chapter 2 — Electrostatic Potential and Capacitance
Next Chapter: Chapter 4 — Moving Charges and Magnetism