# Find Io in the circuit

Find Io in the circuit in Fig. P3.7 using nodal analysis. Image from: J. D. Irwin and R. M. Nelms, Basic engineering circuit analysis, 10th ed. Hoboken, NJ: John Wiley, 2011.

#### Solution:

Let us label the nodes we will use to perform nodal analysis as follows: In the following equations, $k=10^3$ and $m=10^{-3}$

We will now write a KCL equation at node $V_1$, the orange node:

$I_1+I_2+4m=0$

Let us express these currents in terms of voltage and resistance using $I=\dfrac{V}{R}$.

$\dfrac{V_1}{1k}+\dfrac{V_1-V_2}{2k}+4m=0\,\,\,\color{orange} {\text{(eq.1)}}$

Now, we shift our focus to the second node, $V_2$ (purple node).

Again, we will write a KCL equation.

$I_3+I_4=6m$

Expressing the currents in terms of voltage and resistance gives us:

$\dfrac{V_2-V_1}{2k}+\dfrac{V_2}{2k}=6m\,\,\,\color{purple} {\text{(eq.2)}}$

Solving equations 1 and 2 simultaneously gives us :

$V_1=-0.8$ v

$V_2=5.6$ v

We know $I_0=I_2$. Writing $I_2$ in terms of voltage and resistance gives us:

$I_0=\dfrac{V_1-V_2}{2k}$

$I_0=\dfrac{-0.8-5.6}{2k}$

$I_0=\dfrac{-6.4}{2k}$

$I_0=-3.2$ mA

(A negative current value means the direction of the current flow is opposite to the direction shown in the circuit diagram.)

$I_0=-3.2$ mA