Electric Current — Practice Quiz

10 Ω

Step 1: The Wheatstone bridge balance condition is P/Q = R/S. Step 2: Rearranging for S: S = QR/P. Step 3: Substitute: S = (5 × 20) / 10 = 100 / 10 = 10 Ω. Step 4: Verify: P/Q = 10/5 = 2; R/S = 20/10 = 2 ✓ — bridge is balanced. Why 40 Ω is wrong: This comes from S = PR/Q = (10 × 20)/5 = 40, which inverts the formula. 20 Ω simply equals R without applying the balance condition.

8 V

Step 1: Total resistance in the circuit = R + r = 8 + 2 = 10 Ω. Step 2: Current I = E / (R + r) = 10 / 10 = 1 A. Step 3: Terminal voltage V = E - Ir = 10 - (1 × 2) = 10 - 2 = 8 V. Alternatively, V = IR = 1 × 8 = 8 V ✓. Why 10 V is wrong: That would be the EMF, which equals terminal voltage only when no current flows (open circuit). 2 V is the voltage drop across internal resistance only. 5 V has no valid basis here.

Charge (or current)

Step 1: Kirchhoff's First Rule states that the sum of currents entering a junction equals the sum of currents leaving it: ΣI_in = ΣI_out. Step 2: This is based on the principle of conservation of charge — charge cannot accumulate at a junction in a steady-state circuit. Step 3: Energy conservation is expressed through Kirchhoff's Second Rule (Loop Rule), not the Junction Rule. Step 4: Voltage and power are not the conserved quantities at a junction — current is.

0.5 A

Step 1: Power formula: P = VI, so I = P/V. Step 2: Substitute: I = 100 W / 200 V = 0.5 A. Step 3: Units check — W/V = (J/s)/(J/C) = C/s = A ✓. Why 2 A is wrong: That would mean P = 2 × 200 = 400 W, not 100 W. It comes from inverting (V/P instead of P/V). 20 A and 0.05 A are calculation errors with incorrect power-of-10 handling.