Find the node voltages in the circuit in Fig. E3.4.Figure E3.4

Apply Kirchhoff's current law at node \(V_{2}\) in the considered circuit and solve.

\(\frac{V_{2}-V_{1}}{10 \mathrm{k}}+2 I_{0}+\frac{V_{2}}{10 \mathrm{k}}=0\)

\(\frac{V_{2}}{10 \mathrm{k}}-\frac{V_{1}}{10 \mathrm{k}}+2\left(\frac{V_{1}}{10 \mathrm{k}}\right)+\frac{V_{2}}{10 \mathrm{k}}=0\)

\(\frac{V_{2}+V_{2}}{10 \mathrm{k}}=\frac{-2 V_{1}+V_{1}}{10 \mathrm{k}}\)

\(2 V_{2}=-V_{1} \ldots \ldots\) (2)

Substitute \(V_{1}=-2 V_{2}\) in equation (1) and solve for \(V_{2}\).

\(V_{2}=2\left(-2 V_{2}\right)-40\)

\(V_{2}=-4 V_{2}-40\)

\(5 V_{2}=-40\)

\(V_{2}=-8 \mathrm{~V}\)

Therefore, the value of voltage \(V_{2}\) is \(-8 \mathrm{~V}\).

Substitute \(V_{2}=-8 \mathrm{~V}\) in equation \((2)\) and solve for \(V_{1}\).

$$ \begin{aligned} V_{1} &=-2 V_{2} \\ &=(-2)(-8 \mathrm{~V}) \\ &=16 \mathrm{~V} \end{aligned} $$

Therefore, the value of voltage \(V_{1}\) is \(16 \mathrm{~V}\).