Current Electricity — Important Questions

10 Ω

Step 1: Terminal voltage V = ε - Ir, so 10 = 12 - I × 2. Step 2: 2I = 2, so I = 1 A. Step 3: Terminal voltage = voltage across external resistance: V = IR. Step 4: 10 = 1 × R. Step 5: R = 10 Ω. Option A (5 Ω) results from using wrong current. Option C (6 Ω) results from an error in applying Kirchhoff's law. The physical principle: terminal voltage is always less than emf because of voltage drop across internal resistance.

0.68 A

Step 1: For n cells in parallel with same emf ε and internal resistance r: equivalent emf = ε = 1.5 V, equivalent internal resistance r_eq = r/n = 0.5/4 = 0.125 Ω. Step 2: Total resistance = R + r_eq = 2 + 0.125 = 2.125 Ω. Step 3: Total current I = ε / (R + r_eq) = 1.5 / 2.125 = 0.706 ≈ 0.68 A. Step 4: Note that emf does not multiply in parallel — only current capacity increases. Option B (2.73 A) incorrectly multiplies emf. Option C uses wrong formula. The key concept: parallel cells give same voltage but lower internal resistance.

40 cm

Step 1: With P in left gap: P/Q = l₁/(100-l₁) = 60/40 = 3/2. So P = (3/2) × 10 = 15 Ω. Step 2: When P and Q are interchanged, Q is in left gap (10 Ω) and P in right gap (15 Ω). Step 3: Now Q/P = l₂/(100-l₂). So 10/15 = l₂/(100-l₂). Step 4: 2/3 = l₂/(100-l₂). Cross multiply: 200 - 2l₂ = 3l₂, so 5l₂ = 200, l₂ = 40 cm. Option B assumes balance point doesn't change. Option D makes a ratio error. This tests whether students understand the meter bridge formula deeply.

120 cm

Step 1: In a potentiometer, emf is proportional to balancing length: ε ∝ l. Step 2: ε₁ ∝ 75 cm and ε₂ ∝ 45 cm. Step 3: When cells are connected in series aiding, total emf = ε₁ + ε₂. Step 4: So new balancing length = l₁ + l₂ = 75 + 45 = 120 cm. Step 5: This uses the linearity principle of the potentiometer. Option B (30 cm) is for series opposing. Option C and D arise from incorrect averaging. The potentiometer's linear voltage gradient means emfs directly add as lengths.