Find Vo and V1 in Fig. E2.23.Figure E2.23

Apply Kirchhoff current law at node \(a\).

\(I_{1}=I_{2}+I_{5}+I_{6}\)

\(\frac{0-V_{a}}{4 \mathrm{k} \Omega}=\frac{V_{a}-V_{b}}{10 \mathrm{k} \Omega}+\frac{V_{a}-0}{6 \mathrm{k} \Omega}+\frac{V_{a}-0}{12 \mathrm{k} \Omega}\)

\(\frac{V_{a}}{4 \mathrm{k} \Omega}+\frac{V_{a}-V_{b}}{10 \mathrm{k} \Omega}+\frac{V_{a}}{6 \mathrm{k} \Omega}+\frac{V_{a}}{12 \mathrm{k} \Omega}=0\)

\(\frac{V_{a}}{4 \mathrm{k} \Omega}+\frac{V_{a}}{10 \mathrm{k} \Omega}+\frac{V_{a}}{6 \mathrm{k} \Omega}+\frac{V_{a}}{12 \mathrm{k} \Omega}=\frac{V_{b}}{10 \mathrm{k} \Omega}\)

\(V_{a}\left(\frac{1}{4 \mathrm{k} \Omega}+\frac{1}{10 \mathrm{k} \Omega}+\frac{1}{6 \mathrm{k} \Omega}+\frac{1}{12 \mathrm{k} \Omega}\right)=\frac{V_{b}}{10 \mathrm{k} \Omega}\)

\(V_{a}\left(\frac{15+6+10+5}{60 \mathrm{k} \Omega}\right)=\frac{V_{b}}{10 \mathrm{k} \Omega}\)

\(V_{a}(15+6+10+5)=\left(\frac{60 \mathrm{k} \Omega}{10 \mathrm{k} \Omega}\right) V_{b}\)

\(36 V_{a}=6 V_{b}\)

\(V_{b}=6 V_{a}\)

Apply Kirchhoff current law at node b.

\(I_{2}=I_{3}+I_{4}\)

\(\frac{V_{a}-V_{b}}{10 \mathrm{k} \Omega}=20 \mathrm{~mA}+\frac{V_{b}-0}{4 \mathrm{k} \Omega} \quad\left(\right.\) since, \(\left.I_{3}=20 \mathrm{~mA}\right)\)

\(\frac{V_{a}-V_{b}}{10 \mathrm{k} \Omega}-\frac{V_{b}}{4 \mathrm{k} \Omega}=20 \mathrm{~mA}\)

\(\frac{2 V_{a}-2 V_{b}-5 V_{b}}{20 \mathrm{k} \Omega}=20 \mathrm{~mA}\)

\(2 V_{a}-2 V_{b}-5 V_{b}=(20 \mathrm{~mA})(20 \mathrm{k} \Omega)\)

\(2 V_{a}-7 V_{b}=400\)

Substitute the expression of \(V_{b}\).

\(2 V_{a}-7 V_{b}=400\)

\(2 V_{a}-7\left(6 V_{a}\right)=400\)

\(2 V_{a}-42 V_{a}=400\)

\(-40 V_{a}=400\)

\(V_{a}=\frac{400}{-40}\)

\(=-10 \mathrm{~V}\)

Substitute the value of \(V_{a}\) in the equation of \(V_{b}\).

\(V_{b}=6 V_{a}\)

\(=6(-10 \mathrm{~V})\)

\(=-60 \mathrm{~V}\)

Find the voltages represented on the circuit.

\(\begin{aligned} V_{o} &=V_{b} \\ &=-60 \mathrm{~V} \\ V_{1} &=0-V_{a} \\ &=0-(-10 \mathrm{~V}) \\ &=10 \mathrm{~V} \end{aligned}\)

Thus, the voltages \(V_{o}\) and \(V_{1}\) are \(-60 \mathrm{~V}\) and \(10 \mathrm{~V}\) respectively.