Problem — equivalent resistance, series-parallel

Problem

Three resistors $R_1 = 6\ \Omega$, $R_2 = 3\ \Omega$, $R_3 = 2\ \Omega$ are connected to an ideal $12\ \text{V}$ battery: $R_1$ is in series with the parallel combination of $R_2$ and $R_3$. Calculate the equivalent resistance, the current supplied by the battery and the current through $R_2$.

Diagram: $R_1$ in series with the parallel combination $R_2\parallel R_3$, powered by the battery.

Solution

Parallel of $R_2$ and $R_3$: $R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{3 \cdot 2}{3 + 2} = \frac{6}{5} = 1{,}2\ \Omega$

Equivalent resistance (series with $R_1$): $R_\text{eq} = R_1 + R_{23} = 6 + 1{,}2 = 7{,}2\ \Omega$

Current supplied by the battery: $I = \frac{\mathcal{E}}{R_\text{eq}} = \frac{12\ \text{V}}{7{,}2\ \Omega} \approx 1{,}67\ \text{A}$

Current in $R_2$: the voltage across the parallel combination is $V_{23} = I\,R_{23} = 1{,}67 \cdot 1{,}2 = 2{,}0\ \text{V}$ hence $I_2 = \frac{V_{23}}{R_2} = \frac{2{,}0}{3} \approx 0{,}67\ \text{A}$

$\ev{R_\text{eq} = 7{,}2\ \Omega \qquad I \approx 1{,}67\ \text{A} \qquad I_2 \approx 0{,}67\ \text{A}}$

Links

Topics: Electric circuits Concepts: Resistors in series and parallel Skills: Network simplification