A battery of e.m.f. 18 V and internal resistance 2 Ω is connected to a 6 Ω and a 12 Ω resistor in parallel. Find the current from the battery and the terminal p.d.
- Parallel pair: R = (6 × 12) ÷ (6 + 12) = 72 ÷ 18 = 4 Ω.
- Total resistance = 4 + 2 = 6 Ω, so I = 18 ÷ 6 = 3 A.
- Terminal p.d. = E − Ir = 18 − 3 × 2 = 12 V.
- Check: 12 ÷ 6 = 2 A and 12 ÷ 12 = 1 A, which add to 3 A.
Answer: 3 A; terminal p.d. 12 V.