An $18,\text{V}$ ideal battery with internal resistance $r=0.4,\Omega$ is connected to two resistors $R_1=5,\Omega$ and $R_2=15,\Omega$ in parallel. The combination is then connected in series with a resistor $R_3=10,\Omega$. (i) Find the total resistance seen by the battery. (ii) Find the current supplied by the battery. (iii) Find the power dissipated in $R_3$.

Step-by-step solution

1. First find the equivalent resistance of $R_1$ and $R_2$ in parallel: $R_{12}=\frac{R_1{R}_2}{R_1+R_2}=\frac{5\times 15}{5+15}=3.75,\Omega$.
2. Add the series resistor $R_3$ to obtain the total resistance: $R_{\text{tot}}=R_{12}+R_3=3.75+10=13.75,\Omega$.
3. Use Ohm’s law to find the battery current: $I=\frac{\mathcal{E}}{R_{\text{tot}}}=\frac{18}{13.75}\approx 1.31,\text{A}$.
4. Use $P=I^{2}R$ to find the power in $R_3$: $P_{R_3}=I^2{R}_3=(1.31{)}^2\times 10\approx 17.1,\text{W}$.

Concept tested: Series‑parallel circuits and power

Published 15 September 2026 · How this solution was prepared and checked