An ideal battery of emf $\mathcal{E}=15\text{V}$ has an internal resistance $r=0.5\Omega$. It is connected to a load resistor $R=10\Omega$. (i) Find the terminal voltage across the load. (ii) Find the power dissipated in the load.

Step-by-step solution

1. The current supplied by the battery is $I=\mathcal{E}/(R+r)$ by applying Ohm's law to the whole circuit.
2. Substitute: $I=15\text{V}/(10+0.5)\Omega =15/10.5\text{A}=1.4286\text{A}$.
3. Terminal voltage across the load is $V_{\text{term}}=IR$.
4. Calculate: $V_{\text{term}}=1.4286\text{A}\times 10\Omega =14.286\text{V}\approx 14.3\text{V}$.
5. Power in the load is $P=I^{2}R$ (or $VI$).
6. Using $I^{2}R$: $P=(1.4286{)}^2\times 10=2.0408\times 10\text{W}=20.41\text{W}\approx 20.5\text{W}$.

Concept tested: Internal resistance effect

Published 13 September 2026 · How this solution was prepared and checked