Use nodal analysis to find Io in the circuit in Fig. E3.11.Figure E3.11

Write the Kirchhoff's current law equation at super node.

$$ \begin{aligned} &-4 \mathrm{~m}+\frac{V_{1}}{2 \mathrm{k}}+2 \mathrm{~m}+\frac{V_{2}}{2 \mathrm{k}}=0 \\ &-2 \mathrm{~m}+\frac{V_{1}}{2 \mathrm{k}}+\frac{V_{2}}{2 \mathrm{k}}=0 \\ &\frac{-(2 \mathrm{~m})(2 \mathrm{k})+V_{1}+V_{2}}{2 \mathrm{k}}=0 \\ &-4+V_{1}+V_{2}=0 \\ &V_{1}+V_{2}=4 \ldots \ldots \text { (3) } \end{aligned} $$

Substitute \(2 V_{1}\) for \(V_{2}\) in equation (3).

$$ \begin{aligned} &V_{1}+2 V_{1}=4 \\ &3 V_{1}=4 \\ &V_{1}=\frac{4}{3} \mathrm{~V} \end{aligned} $$

Substitute \(\frac{4}{3} \mathrm{~V}\) for \(V_{1}\) in equation (2).

$$ \begin{aligned} V_{2} &=2\left(\frac{4}{3}\right) \\ &=\frac{8}{3} \mathrm{~V} \end{aligned} $$

From Figure E3.11, the current \(I_{o}\) is,

$$ \begin{aligned} I_{e} &=\frac{V_{2}}{2 \mathrm{k}} \\ &=\frac{\frac{8}{3}}{2 \mathrm{k}} \\ &=\frac{4}{3} \mathrm{~mA} \end{aligned} $$

Therefore, the value of current, \(I_{o}\) is \(\frac{4}{3} \mathrm{~mA}\)