# Find Io in the circuit in Fig. P3.10

Find Io in the circuit in Fig. P3.10 using nodal analysis.

Image from: J. D. Irwin and R. M. Nelms, Basic engineering circuit analysis, 10th ed. Hoboken, NJ: John Wiley, 2011.

#### Solution:

To see the nodes clearly, we will “bend” the wires as follows:

Note that this does not change the original circuit diagram, rather, gives us a new perspective and makes it easier to work with. This can be done as long as there are “empty” wire segments connecting circuit elements together.

Let us now label our nodes and currents. Remember that we must also have a reference ground node, which we chose to be right side of the circuit as shown below.

In the following equations, $k=10^3$ and $m=10^{-3}$

Looking at the orange node, $V_1$, we will write a KCL equation.

$I_1+I_2+2m=1m$

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

$\dfrac{V_1-V_2}{2k}+\dfrac{V_1}{8k}+2m=1m\,\,\,\color{orange} {\text{(eq.1)}}$

Now, we will shift our attention to the purple node, $V_2$ and write a KCL equation.

$I_3+I_4+I_5=2m$

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

$\dfrac{V_2}{6k}+\dfrac{V_2-V_1}{2k}+\dfrac{V_2}{3k}=2m\,\,\,\color{purple} {\text{(eq.2)}}$

Solving equations 1 and 2 simultaneously gives us :

$V_1=0$ v

$V_2=2$ v

From our diagram, we know that:

$I_0=I_3$

$I_0=\dfrac{V_2}{6k}$

$I_0=\dfrac{2}{6k}$

$I_0=0.33$ mA

$I_0=0.33$ mA