# Use nodal analysis to find both V1 and Vo

Use nodal analysis to find both V1 and Vo in the circuit in Fig 3.6.

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 currents flowing from each node as follows:

When writing equations, $k=10^3$ and $m=10^{-3}$

Note that we will use $I_4$ to calculate $V_0$.

Let us write a KCL equation for node $V_1$, the orange node.

$I_1+I_2+2m=12m$

We can express these currents in terms of voltage and resistance using I=V/R.

$\dfrac{V_1}{3k}+\dfrac{V_1-V_2}{6k}+2m=12m\,\,\,\color{orange} {\text{(eq.1)}}$

We will now look at node $V_2$, which is the purple node and write a KCL equation.

$I_3+I_4+I_5=2m$

Again, we will express these currents in terms of voltage and resistance.

$\dfrac{V_2-V_1}{6k}+\dfrac{V_2}{2k+1k}+\dfrac{V_2}{6k}=2m\,\,\,\color{purple} {\text{(eq.2)}}$

(Remember that for $I_4$, the current that flows is across both resistors added together as they are in series)

Solving equations 1 and 2 simultaneously gives us (full steps here):

$V_1=\dfrac{252}{11}$ v

$V_2=\dfrac{96}{11}$ v

We will now use the value of $V_2$ to find $V_0$. Remember that

$I_4=\dfrac{V_2}{2k+1k}$

$I_4=\dfrac{(\frac{96}{11})}{3k}$

$I_4=2.91$ mA

Now,

$V_0=(I_4)(1k)$

$V_0=(2.91m)(1k)$

$V_0=2.91$ v

$V_1=\dfrac{252}{11}$ v
$V_0=2.91$ v