Current Electricity Class 12 Notes - Chapter 3 Short Notes

Chapter summary

Current Electricity (Class 12 Physics) is about how electric charge flows through conductors and the laws that govern it. Electric current is the rate of flow of charge, I = Q/t, measured in amperes. Ohm’s law states V = IR, where resistance R = rho L/A depends on the material and dimensions of the conductor. Drift velocity, resistivity, internal resistance of a cell, and Kirchhoff’s two rules (junction and loop) let you solve any circuit, and a Wheatstone bridge or potentiometer measures unknown resistances and EMFs precisely. Key formulas: I = nAev_d, V = IR, P = VI = I^2 R = V^2/R, and EMF = V + Ir.

Chapter notes

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)


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)


3. Combinations of Resistors

FeatureSeriesParallel
CurrentSameDivides
VoltageDividesSame
Equivalent RR = R₁ + R₂ + …1/R = 1/R₁ + 1/R₂ + …

SeriesR₁R₂Rₛ = R₁ + R₂ (same current)ParallelR₁R₂1/Rₚ = 1/R₁ + 1/R₂ (same voltage)
Resistors in series add directly; in parallel the reciprocals add. Series carries the same current, parallel shares the same voltage.

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

TypeEMFInternal ResistanceBest for
Series (n cells)nrHigh EMF needed (external R >> internal r)
Parallel (n cells)εr/nMore current needed (external R << internal r)

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


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

PQRSGABCD
Wheatstone bridge. At balance the galvanometer G reads zero and P/Q = R/S, so an unknown resistance can be found from the other three.

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

Weightage in Board & Entrance Exams

ExamTypical WeightageMost-Tested Areas
CBSE Board (Class 12)7–9 marksOhm's law, resistor networks, Kirchhoff's laws, potentiometer, Wheatstone bridge
JEE Main / Advanced2–3 questionsEquivalent resistance, cells in series/parallel, meter bridge, RC ideas
NEET2–3 questionsDrift velocity, resistivity, Wheatstone balance, internal resistance

Important Definitions

TermDefinition
Electric currentRate of flow of charge: I = dQ/dt
Drift velocityAverage velocity of free electrons in a conductor under applied field
ResistivityMaterial property that determines resistance: R = ρl/A
EMFWork done per unit charge by the cell in moving charge through the complete circuit
Internal resistanceResistance offered by the electrolyte and electrodes inside the cell
Kirchhoff’s junction ruleSum of currents at a junction = 0 (conservation of charge)
Kirchhoff’s loop ruleSum of potential differences in a closed loop = 0
Wheatstone bridgeCircuit of four resistors; balanced when P/Q = R/S

Read the rest of the chapter →Hide the rest ↑

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.


Example 5

Q. In a Wheatstone bridge P = 10 Ω, Q = 20 Ω and R = 15 Ω. Find S for a balanced bridge.

A. At balance P/Q = R/S, so S = R×Q/P = 15×20/10 = 30 Ω.

Example 6

Q. Two resistors 6 Ω and 3 Ω are connected in parallel. What is their equivalent resistance?

A. 1/Rₚ = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, so Rₚ = 2 Ω (always less than the smallest resistor).

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?

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)

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

🃏 Flash Cards: Current Electricity

Class 12 Physics · Chapter 3 – swipe through all 10 cards to understand the whole chapter.

🔌Start here1/10

Electric Current

Rate of flow of charge through a cross-section of a conductor.

I = q / t

Unit: ampere (A) · scalar

  • Direction = flow of +ve charge
  • Steady current: constant I
  • 1 A = 1 C/s
🌊Microscopic view2/10

Drift Velocity

Average velocity electrons gain along the wire under a field.

I = n e A v_d

v_d = (eE/m)·τ

  • Tiny (~10⁻4 m/s)
  • n = electrons per unit volume
  • τ = relaxation time
📏Core law3/10

Ohm’s Law & Resistance

Current is proportional to voltage for an ohmic conductor.

V = I R , R = ρL / A

ρ = resistivity (Ω·m)

  • R rises with length L
  • R falls with area A
  • Unit of R: ohm (Ω)
🌡️Material effect4/10

Resistivity & Temperature

Resistivity of a metal increases with temperature.

ρ_T = ρ0 (1 + α ΔT)

α = temperature coefficient

  • Metals: α positive
  • Semiconductors: α negative
  • Superconductors: ρ → 0
🪜Networks5/10

Combination of Resistors

How resistance adds when resistors are joined.

Series: R = Σ Rᵢ · Parallel: 1/R = Σ 1/Rᵢ

Series same I · Parallel same V

  • Series R > largest
  • Parallel R < smallest
  • Opposite of capacitors
🔋Real cells6/10

EMF & Internal Resistance

A real cell loses some voltage across its own resistance r.

V = E − I r , I = E / (R + r)

E = emf, V = terminal voltage

  • V < E while discharging
  • V = E on open circuit
  • Short circuit I = E/r
⚖️Circuit rules7/10

Kirchhoff’s Laws

Two conservation rules to solve any circuit network.

KCL: Σ I = 0 · KVL: Σ V = 0

Junction (charge) · Loop (energy)

  • KCL: ΣI in = ΣI out
  • KVL: ΣEMF = ΣIR in a loop
  • Sign convention matters
🌉Measurement8/10

Wheatstone Bridge

A network to measure an unknown resistance at balance.

Balanced: P / Q = R / S

Galvanometer reads zero

  • No current through bridge arm
  • Basis of the meter bridge
  • Most sensitive when arms equal
📐Compare EMFs9/10

Potentiometer

Measures potential difference without drawing current.

V ∝ l → E1 / E2 = l1 / l2

Balancing length l

  • More accurate than a voltmeter
  • Draws zero current at balance
  • Compares EMFs of two cells
💡Energy10/10

Electric Power

Rate at which electrical energy is converted in a device.

P = V I = I2 R = V2 / R

Unit: watt (W) · H = I2Rt

  • Heating: Joule’s law
  • Bulb rating = P at rated V
  • 1 kWh = 1 ‘unit’
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📝 Practice Current Electricity — 10 NEET PYQs
Real previous-year questions · with answers & solutions
Start →Close ✕
Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2021
In a potentiometer circuit, a cell of EMF 1.5 V gives a balance point at 36 cm length of wire. If a cell of EMF 2.5 V replaces the first cell, the balance point occurs at:
Correct answer: A. With a constant potential gradient, the balancing length is proportional to EMF: (L₁)/(L₂) = (ε₁)/(ε₂)L₂ = 36×(2.5)/(1.5) = 60 cm.
🔎 See the full step-by-step solution in the app →
Q2NEET 2020
The colour code on a carbon resistor is Yellow, Violet, Brown, Gold (in order). Its resistance and tolerance are:
Correct answer: C. Carbon-resistor code: Yellow = 4, Violet = 7, so the first two digits give 47. Brown is the multiplier 10¹, giving 47×10¹ = 470 Ω. Gold tolerance band = ±5%. So the resistor is 470 Ω, 5%.
🔎 See the full step-by-step solution in the app →
Q3NEET 2019
Which of the following acts as a circuit-protection device?
Correct answer: C. A fuse is a short piece of wire of high resistivity and low melting point. When the current exceeds a safe value (e.g. a short circuit), Joule heating melts the fuse and breaks the circuit, protecting the appliances.
🔎 See the full step-by-step solution in the app →
Q4NEET 2017
The resistance of a wire is R ohm. If it is melted and stretched to n times its original length, its new resistance is:
Correct answer: C. Melting and redrawing conserves volume, so when length becomes nL the area becomes A/n. Since R = ρ L/A, the new resistance R’ = ρ(nL)/(A/n) = n² ρ(L)/(A) = n²R.
🔎 See the full step-by-step solution in the app →
Q5NEET 2015
Two metal wires of identical dimensions are connected in series. If σ₁ and σ₂ are their conductivities, the effective conductivity of the combination is:
Correct answer: A. In series R_(eq)=R₁+R₂. With identical dimensions, ρ_(eff)(2L)/(A)=ρ₁(L)/(A)+ρ₂(L)/(A), so 2ρ_(eff)=ρ₁+ρ₂. Using σ=1/ρ: (2)/(σ_(eff))=(1)/(σ₁)+(1)/(σ₂), giving σ_(eff)=(2σ₁σ₂)/(σ₁+σ₂).
🔎 See the full step-by-step solution in the app →
Q6NEET 2013
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is:
Correct answer: B. ε = I(R + r)2.1 = 0.2(10 + r)10 + r = 10.5r = 0.5 Ω.
🔎 See the full step-by-step solution in the app →
Q7NEET 2012
If the voltage across a bulb rated ‘220 V, 100 W’ drops by 2.5% of its rated value, the percentage by which the power decreases is (assume constant resistance):
Correct answer: C. Since P = V²/R with R constant, (Δ P)/(P) = 2(Δ V)/(V). For a 2.5% drop in V, the power decreases by 2×2.5% = 5%.
🔎 See the full step-by-step solution in the app →
Q8NEET 2011
A current of 2 A flows through a 2 Ω resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 Ω resistor. The internal resistance of the battery is:
Correct answer: A. ε = I(R + r). Case 1: ε = 2(2 + r) = 4 + 2r. Case 2: ε = 0.5(9 + r) = 4.5 + 0.5r. Equating: 4 + 2r = 4.5 + 0.5r1.5r = 0.5r = (1)/(3) Ω.
🔎 See the full step-by-step solution in the app →
Q9NEET 2004
n resistors, each of r ohm, when connected in parallel give an equivalent resistance of R ohm. If these resistances are instead connected in series, the combination’s resistance equals:
Correct answer: A. In parallel, R = r/n, so r = nR. In series, Rₛ = nr = n(nR) = n²R.
🔎 See the full step-by-step solution in the app →
Q10NEET 1995
Two wires of the same metal have the same length, but their cross-sections are in the ratio 3 : 1. They are joined in series. If the resistance of the thicker wire is 10 Ω, the total resistance of the combination is:
Correct answer: C. For the same material and length, R ∝ 1/A. The thin wire has one-third the area, so its resistance is 3×10 = 30 Ω. In series the total is 10 + 30 = 40 Ω.
🔎 See the full step-by-step solution in the app →
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Frequently Asked Questions

What are the most important topics in Current Electricity Class 12?

Ohm’s law and resistance, series and parallel combinations of resistors, cells and internal resistance, Kirchhoff’s laws, the Wheatstone bridge, the meter bridge and the potentiometer. Drift velocity and resistivity-temperature dependence are frequent 1-mark and assertion-reason questions.

Is Current Electricity important for NEET and JEE?

Yes - it is one of the highest-weightage Class 12 Physics chapters in both exams. Numericals on Kirchhoff’s laws, internal resistance and resistor combinations appear almost every year.

Do these notes include important questions for board exams?

Yes. The notes end with 1-mark, 3-mark and 5-mark important questions, four solved examples and quick revision points covering the formulas CBSE asks most often.

Are these notes based on the NCERT syllabus?

Yes - they follow NCERT Class 12 Physics Chapter 3 (Current Electricity) as per the current CBSE syllabus, written as short notes for quick revision.

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