A cell of emf 9.0 V and internal resistance 1.0 Ω is connected to two 10 Ω resistors that are in parallel with each other. What power is dissipated in each resistor?
- External resistance: 1/R = 1/10 + 1/10, so R = 5.0 Ω.
- Current from the cell: I = ε ÷ (R + r) = 9.0 ÷ (5.0 + 1.0) = 1.5 A.
- Terminal potential difference: V = ε − Ir = 9.0 − 1.5 × 1.0 = 7.5 V, and this is across each resistor.
- Power in each: P = V2 ÷ R = 7.52 ÷ 10 = 56.25 ÷ 10 = 5.6 W.
Answer: About 5.6 W in each resistor.