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Current Electricity – Class 12 Physics notes, important questions and MCQs

NCERT Class 12 Physics, chapter 3. Notes, key terms, important questions with answers and MCQs.

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Chapter overview

Current is the rate of flow of charge. Ohm's law, resistivity, cells and Kirchhoff's rules analyse circuits.

Current and resistance

I = q/t = neAv_d (drift velocity v_d).
Ohm's law V = IR; R = ρl/A; resistivity increases with temperature for metals.
Series R = R₁ + R₂; parallel 1/R = 1/R₁ + 1/R₂.

Cells and networks

Terminal voltage V = E − Ir (discharging); r is internal resistance.
Kirchhoff's junction rule (charge conservation) and loop rule (energy conservation).
Wheatstone bridge balanced when P/Q = R/S; power P = VI = I²R.

Key terms

Drift velocity: average velocity of electrons in a conductor under an electric field
Internal resistance: resistance offered by the cell's electrolyte
Ohm's law: at constant temperature, current through a conductor is proportional to the potential difference, V = IR

Formulas

V = IR
R = ρl/A
V = E − Ir
P/Q = R/S (balanced bridge)

Important questions with answers

Define drift velocity and relate it to current.2 marks

Drift velocity is the average velocity with which free electrons move opposite to the field in a conductor. I = neAv_d, where n is the free-electron density and A the cross-sectional area.

State Kirchhoff's two rules and the conservation law each is based on.3 marks

Junction rule: the sum of currents entering a junction equals the sum leaving (conservation of charge). Loop rule: the algebraic sum of potential changes around a closed loop is zero (conservation of energy).

Derive the balance condition of a Wheatstone bridge. In a metre bridge the null point is at 40 cm with a 6 Ω resistance in the left gap; find the unknown resistance in the right gap.5 marks

Bridge: P, Q, R, S with a galvanometer between the junctions of P–Q and R–S. At balance Ig = 0, so the same current I₁ flows in P and Q, I₂ in R and S, and the two junctions are at the same potential: I₁P = I₂R and I₁Q = I₂S. Dividing: P/Q = R/S. Metre bridge: R/S = l/(100 − l), so 6/S = 40/60 and S = 9 Ω.

Quick revision

Junction rule: ΣI = 0.
Loop rule: ΣV = 0.
Power P = I²R.

Exam tips

In a balanced Wheatstone bridge, no current flows through the galvanometer.
In Kirchhoff problems, choose current directions freely; a negative answer only means the actual direction is opposite.

Practice MCQs

SI unit of resistivity is
1. ohm
2. ohm-metre
3. ohm/metre
4. siemens
Show answer
Answer: (b) ohm-metre. Ω m.
Kirchhoff's junction rule is based on conservation of
1. energy
2. charge
3. momentum
4. mass
Show answer
Answer: (b) charge. Charge.
A wire of resistance 10 Ω is stretched to twice its length. New resistance is
1. 20 Ω
2. 40 Ω
3. 5 Ω
4. 10 Ω
Show answer
Answer: (b) 40 Ω. R ∝ l².
A cell of emf 2 V and internal resistance 0.5 Ω gives 1 A. Terminal voltage is
1. 2 V
2. 1.5 V
3. 2.5 V
4. 1 V
Show answer
Answer: (b) 1.5 V. 2 − 1 × 0.5.
A wire of resistance 10 Ω is stretched to twice its length. New resistance is
1. 20 Ω
2. 40 Ω
3. 5 Ω
4. 10 Ω
Show answer
Answer: (b) 40 Ω. R ∝ l² at constant volume.
In a balanced Wheatstone bridge P = 2 Ω, Q = 4 Ω, R = 6 Ω. S is
1. 3 Ω
2. 8 Ω
3. 12 Ω
4. 6 Ω
Show answer
Answer: (c) 12 Ω. 2/4 = 6/S.
Resistivity of a metal on heating
1. decreases
2. stays the same
3. increases
4. becomes zero
Show answer
Answer: (c) increases. More collisions.
A cell of emf 2 V and internal resistance 0.5 Ω is connected to 3.5 Ω. Terminal voltage is
1. 2 V
2. 1.75 V
3. 1.5 V
4. 0.25 V
Show answer
Answer: (b) 1.75 V. I = 0.5 A; V = 2 − 0.25.

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Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.