Question

Figure (32-E10) shows a part of an electric circuit. The potentials at the points a, b and c are 30 V, 12 V and 2 V respectively. Find the currents through the three resistors.

Answer 1

Let the potential at the point o be X volts.
From the figure,
$$\begin{gathered} i_1=\frac{V_{\mathrm{a}}-V_{\mathrm{o}}}{10} \\ {V}_{\mathrm{a}}=30\mathrm{V}\mathrm{and}{V}_{\mathrm{o}}=X \\ So,i_1=\frac{30-X}{10} \\ \mathrm{Similarly}, \\ {i}_2=\frac{V_{\mathrm{o}}-V_{\mathrm{b}}}{20} \\ =\frac{X-12}{20} \\ \mathrm{And} \\ {i}_3=\frac{V_{\mathrm{o}}-V_{\mathrm{c}}}{30} \\ =\frac{X-2}{30} \end{gathered}$$
Also, from kirchoff's junction law we have:
i₁ = i₂ + i₃
$$\begin{gathered} \Rightarrow \frac{30-X}{10}=\frac{X-12}{20}+\frac{X-2}{30} \\ \Rightarrow 30-X=\frac{X-12}{2}+\frac{X-2}{3} \\ \Rightarrow 30-X=\frac{3X-36+2X-4}{6} \\ \Rightarrow 180-6X=5X-40 \\ \Rightarrow 11X=220 \\ \Rightarrow X=\frac{220}{11}=20\mathrm{V} \end{gathered}$$
Thus, the currents through the three resistors are:

$$\begin{gathered} i_1=\frac{30-20}{10}=1\mathrm{A} \\ {i}_2=\frac{20-12}{20}=\frac{8}{20}=0.4\mathrm{A} \\ {i}_3=\frac{20-2}{30}=\frac{18}{30}=0.6\mathrm{A} \end{gathered}$$