A $9\text{V}$ ideal battery has an internal resistance of $0.2\Omega$. It is connected to a lamp whose resistance is $3\Omega$. (i) Find the terminal voltage across the lamp. (ii) Find the power dissipated in the lamp.

Step-by-step solution

1. The total resistance in the circuit is $R_{\text{total}}=R_{\text{int}}+R_{\text{lamp}}=0.2\Omega +3\Omega =3.2\Omega$.
2. Current supplied by the battery: $I=\frac{\mathcal{E}}{R_{\text{total}}}=\frac{9\text{V}}{3.2\Omega }=2.8125\text{A}$.
3. Terminal voltage across the external lamp is $V_{\text{term}}=I\times R_{\text{lamp}}=2.8125\text{A}\times 3\Omega =8.4375\text{V}$.
4. Rounded to one decimal place, $V_{\text{term}}\approx 8.4\text{V}$.
5. Power in the lamp: $P=I^2{R}_{\text{lamp}}=(2.8125{)}^2\times 3=23.5\text{W}$ (rounded to one decimal).

Concept tested: Terminal voltage and power with internal resistance

Published 8 September 2026 · How this solution was prepared and checked