A $12,\text{V}$ ideal battery with internal resistance $r=0.30,\Omega$ is connected to two resistors $R_1=3,\Omega$ and $R_2=6,\Omega$ in parallel. This parallel combination is then connected in series with a resistor $R_3=9,\Omega$. Using Kirchhoff’s laws, find (i) the total current supplied by the battery and (ii) the current through each resistor $R_1,R_2,$ and $R_3$.

Step-by-step solution

1. Find the equivalent resistance of the parallel pair: $R_{12}=(3\times 6)/(3+6)=2,\Omega$.
2. Add $R_{12}$ to $R_3$ for the series resistance: $R_{\text{series}}=2+9=11,\Omega$.
3. Add the internal resistance of the battery: $R_{\text{total}}=11+0.30=11.30,\Omega$.
4. Compute the total current from Ohm’s law: $I_{\text{total}}=V/R_{\text{total}}=12/11.30\approx 1.06,\text{A}$.
5. The voltage across the parallel pair equals $V_{\text{parallel}}=I_{\text{total}}\times R_{12}=1.06\times 2\approx 2.12,\text{V}$.
6. Current through $R_1$: $I_{R1}=V_{\text{parallel}}/R_1=2.12/3\approx 0.71,\text{A}$.
7. Current through $R_2$: $I_{R2}=V_{\text{parallel}}/R_2=2.12/6\approx 0.35,\text{A}$.
8. Since $R_3$ is in series, $I_{R3}=I_{\text{total}}\approx 1.06,\text{A}$.

Concept tested: Kirchhoff’s laws in a mixed circuit

Published 19 September 2026 · How this solution was prepared and checked