Current Electricity
Punjab Board · Class 12 · Physics
Most important questions from Current Electricity for Punjab Board Class 12 Physics board exam 2026. MCQs, short answer, and long answer questions with marks.
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The resistance of a metallic wire is doubled by doubling its length (keeping area constant). If the original resistance is 10 Ω, what is the new resistance?
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20 Ω
Step 1: From R = ρl/A, resistance R is directly proportional to length l (when area A and resistivity ρ are constant). Step 2: If length is doubled (l → 2l), then R also doubles: R_new = 2 × R_original. Step 3: R_new = 2 × 10 = 20 Ω. Option A (5 Ω) would be the case if length was halved. Option B (10 Ω) assumes no change. Option D (40 Ω) would mean resistance quadrupled, which happens if length doubles AND area halves simultaneously.
The EMF of a cell is 12 V and its internal resistance is 2 Ω. What is the terminal voltage when a current of 2 A is drawn from it?
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10 V
Step 1: Recall the formula for terminal voltage: V = ε - Ir, where ε is EMF, I is current, and r is internal resistance. Step 2: Substitute values: ε = 12 V, I = 2 A, r = 2 Ω. Step 3: V = 12 - (2 × 2) = 12 - 4 = 10 V. Option A (12 V) is the EMF, which equals terminal voltage only when no current flows (open circuit). Option C (8 V) would mean Ir = 4 V but with I=4 A. Option D (14 V) would result from adding instead of subtracting — terminal voltage is always less than EMF when current flows.
Which of the following correctly states Kirchhoff's Junction Rule (KCL)?
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The sum of currents entering a junction equals the sum of currents leaving the junction
Step 1: Kirchhoff's rules consist of two rules — the Junction Rule (KCL) and the Loop Rule (KVL). Step 2: The Junction Rule states that the algebraic sum of all currents at a junction is zero, which means total current entering = total current leaving. This is based on conservation of charge — charge does not accumulate at a junction in a steady state. Step 3: Option A describes Kirchhoff's Loop Rule (KVL), not the junction rule. Option C is incorrect — currents in different branches can be different. Option D is wrong — voltage across parallel resistors is the same.
A resistor dissipates 100 W of power when connected to a 200 V supply. What is the resistance of the resistor?
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400 Ω
Step 1: Use the power formula P = V²/R, which gives R = V²/P. Step 2: Substitute values: V = 200 V, P = 100 W. Step 3: R = (200)²/100 = 40000/100 = 400 Ω. Option A (2 Ω) comes from using R = P/V = 100/200 incorrectly. Option B (20 Ω) is obtained by forgetting to square the voltage. Option D (0.5 Ω) comes from inverting the formula. Always remember: P = V²/R means R = V²/P.
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