An ideal battery of emf 12 V and internal resistance 0.5 Ω supplies current to two resistors $R_1=4,Omega$ and $R_2=6,Omega$ connected in parallel. Calculate (i) the current through each resistor, (ii) the total power delivered by the battery. Then a switch is closed connecting $R_2$ in series with a third resistor $R_3=3,Omega$ (parallel with $R_1$). Find the new current from the battery and the new total power dissipated.

Step-by-step solution

1. 1. Equivalent resistance of $R_1$ and $R_2$ in parallel: $R_p=(R_1{R}_2)/(R_1+R_2)=24/10=2.4,Omega$.
2. 2. Total resistance including internal: $R_{\text{tot1}}=0.5+2.4=2.9,\Omega$.
3. 3. Total current from battery: $I_1=V/R_{\text{tot1}}=12/2.9=4.14,\text{A}$.
4. 4. Voltage across the parallel network: $V_p=I_1{R}_p=4.14\times 2.4=9.93,\text{V}$.
5. 5. Current through $R_1$: $I_{R1}=V_p/R_1=9.93/4=2.48,\text{A}$.
6. 6. Current through $R_2$: $I_{R2}=V_p/R_2=9.93/6=1.66,\text{A}$.
7. 7. Power delivered by battery: $P_1=V{I}_1=12\times 4.14=49.7,\text{W}$.
8. 8. After closing switch: $R_2$ and $R_3$ are in series: $R_{23}=6+3=9,Omega$.
9. 9. Equivalent of $R_{23}$ in parallel with $R_1$: $R_{p2}=(R_{23}{R}_1)/(R_{23}+R_1)=36/13=2.77,Omega$.
10. 10. Total resistance with internal: $R_{\text{tot2}}=0.5+2.77=3.27,\Omega$.
11. 11. New total current: $I_2=12/3.27=3.67,\text{A}$.
12. 12. New power delivered: $P_2=12\times 3.67=44.0,\text{W}$.
13. 13. Final results: $I_1=4.14,\text{A}$, $I_{R1}=2.48,\text{A}$, $I_{R2}=1.66,\text{A}$, $P_1=49.7,\text{W}$, $I_2=3.67,\text{A}$, $P_2=44.0,\text{W}$.

Concept tested: Kirchhoff’s circuit laws

Published 3 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked