Problem

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

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

Figure E3.11

#### Step-by-Step Solution

Solution 1

Refer to Figure E3.11 in the text book.

From Figure E3.11, the current $$I_{x}$$ is,

$$I_{x}=\frac{V_{1}}{2 \mathrm{k}} \ldots \ldots(1)$$

From Figure E3.11,

$$V_{2}-V_{1}=2000 I_{x}$$

Substitute $$\frac{V_{1}}{2 \mathrm{k}}$$ for $$I_{x}$$.

\begin{aligned} &V_{2}-V_{1}=2000\left(\frac{V_{1}}{2 \mathrm{k}}\right) \\ &V_{2}-V_{1}=V_{1} \\ &V_{2}=2 V_{1} \ldots \ldots(2) \end{aligned}

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 $$\frac{4}{3} \mathrm{~mA}$$