A 12 V ideal battery has an internal resistance of 0.3 Ω and powers a lamp of resistance 5 Ω. (i) What is the terminal voltage across the lamp? (ii) What is the power dissipated in the lamp?

Step-by-step solution

1. Calculate the total resistance of the circuit: $R_{\text{tot}}=5+0.3=5.3,\Omega$.
2. Use Ohm’s law to find the current: $I=V/R_{\text{tot}}=12/5.3\approx 2.264,\text{A}$.
3. Find the terminal voltage across the lamp: $V_{\text{lamp}}=I\times 5=11.32,\text{V}$.
4. Calculate the power in the lamp using $P=I^{2}R$ (or $V_{\text{lamp}}\times I$): $P=2.264^2\times 5\approx 25.6,\text{W}$.

Concept tested: Battery internal resistance and power dissipation

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked