# Find V0 in the network in Fig. P3.12

Find V0 in the network in Fig. P3.12 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 our nodes and currents as follows: As there are two voltage sources, the blue node and the green node will be kept at a certain voltage difference between the reference node (ground node) and the selected node. In other words, the blue node is 12 v as there is a 12 voltage source supply power, and the green node will be kept at 6 v due to the 6 v power supply directly corrected.

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

$I_1+I_2+I_3=0$

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

$\dfrac{V_1}{6000}+\dfrac{V_1-6}{12000}+\dfrac{V_1-12}{6000}=0$

Solving for $V_1$ gives us:

$V_1=6$ v

From the diagram, we see that $V_0$ is the difference between the orange node ($V_1$) and the green node. Thus, we have:

$V_0=6 - 6$

$V_0=0$ v

$V_0=0$ v