Worked exercise — mixed series-parallel network

Problem

A circuit has a generator $\mathcal{E} = 8\;\text{V}$ and three resistors: $R_1 = 6\;\Omega$, $R_2 = 1\;\Omega$, $R_3 = 7\;\Omega$, with $R_2$ and $R_3$ in parallel with each other and this group in series with $R_1$. Find the total resistance, the current $i_1$ supplied by the generator and the currents $i_2$, $i_3$ in the parallel branch.

$R_1$ in series with the parallel of $R_2$ and $R_3$.

Solution

Step 1 — equivalent of the parallel $R_2 \parallel R_3$. $R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{1 \cdot 7}{1 + 7} = \frac{7}{8} = 0{,}875\;\Omega$

Step 2 — reduced circuit: $R_1$ in series with $R_{23}$. $R_{\text{tot}} = R_1 + R_{23} = 6 + 0{,}875 = 6{,}875\;\Omega$ $i_1 = \frac{\mathcal{E}}{R_{\text{tot}}} = \frac{8}{6{,}875} \approx 1{,}16\;\text{A}$

Step 3 — currents in the parallel branch. The parallel combination has across it the same voltage $V_{23} = R_{23}\,i_1$: $V_{23} = 0{,}875 \cdot 1{,}16 \approx 1{,}02\;\text{V}$ $i_2 = \frac{V_{23}}{R_2} = \frac{1{,}02}{1} \approx 1{,}02\;\text{A} \qquad i_3 = \frac{V_{23}}{R_3} = \frac{1{,}02}{7} \approx 0{,}15\;\text{A}$

Energy check. Power supplied $P = \mathcal{E}\,i_1 = 8 \times 1{,}16 = 9{,}28\;\text{W}$; power dissipated $P_{R_1} + P_{R_2} + P_{R_3} = 8{,}10 + 1{,}03 + 0{,}15 = 9{,}28\;\text{W}$. Balance verified.

$\ev{R_{\text{tot}} = 6{,}875\;\Omega,\quad i_1 \approx 1{,}16\;\text{A},\quad i_2 \approx 1{,}02\;\text{A},\quad i_3 \approx 0{,}15\;\text{A}}$

Links

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