# Use nodal analysis to find V1

Use nodal analysis to find V1 in the circuit in Fig P3.4. Image from: J. D. Irwin and R. M. Nelms, Basic engineering circuit analysis, 10th ed. Hoboken, NJ: John Wiley, 2011.

#### Solution:

Let us first label the two nodes we will use to perform nodal analysis as follows: Both of the selected nodes have been labeled $V_1$ and $V_2$. The currents leaving the nodes are arbitrarily chosen and labeled. It is common to assume that all currents leave the node. In addition, we assume currents leaving the node to be positive, and currents entering the node to be negative, again, this is arbitrarily chosen and can be switched around if you wanted.

(Note that $k=10^3$ and $m=10^{-3}$)

Let us start with node $V_1$ (orange node).
Using KCL, we can sum up all currents leaving the node.

$I_1+I_2+6m=0$

Let us now express these currents in terms of voltage and resistance using (I=V/R).

$\dfrac{V_1}{3k}+\dfrac{V_1-V_2}{2k}+6m=0\,\,\,\color{orange} {\text{(eq.1)}}$

Next, we look at the purple node $V_2$. Again, let us write our KCL expression.

$I_3+I_4+I_5+4m=0$

As before, we will express the currents in terms of voltage and resistance.

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

We can solve equations 1 and 2 simultaneously to get:

$V_1=-11$ v

$V_2=-6.33$ v

(note that we haven’t done anything special to get these answers, we simply solved the two equations using basic algebra. Fully solved steps here)

$V_1=-11$ v

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