Current Electricity: Exam-Pattern Drill (2)

For $R_2 \parallel R_3$:

$R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{6 \times 3}{9} = 2\,\Omega$

Add the internal resistance, $R_1$, and $R_{23}$ in series:

$R_{\text{total}} = r + R_1 + R_{23} = 1 + 4 + 2 = 7\,\Omega$

$I = \frac{\varepsilon}{R_{\text{total}}} = \frac{12}{7} \approx 1.71\,\text{A}$

This current flows through the battery and through $R_1$.

Voltage across the parallel combination:

$V_{23} = I \cdot R_{23} = \frac{12}{7} \times 2 = \frac{24}{7} \approx 3.43\,\text{V}$

Current through each branch:

$I_2 = \frac{V_{23}}{R_2} = \frac{24/7}{6} = \frac{4}{7} \approx 0.57\,\text{A}$

$I_3 = \frac{V_{23}}{R_3} = \frac{24/7}{3} = \frac{8}{7} \approx 1.14\,\text{A}$

Check: $I_2 + I_3 = 12/7 = I$. Kirchhoff’s junction law passes.

$V_{\text{terminal}} = \varepsilon - Ir = 12 - \frac{12}{7} \times 1 = \frac{72}{7} \approx 10.29\,\text{V}$

Final: $I_1 \approx 1.71$ A, $I_2 \approx 0.57$ A, $I_3 \approx 1.14$ A, $V \approx 10.29$ V.