Electronic – Determine the node voltages va and vb for the circuit

circuit analysis

enter image description here

I set up the equations $$ -\frac{V_a}{4\Omega} – 3A – \frac{V_a – V_b}{2\Omega} = 0 \\\frac{V_a – V_b}{2\Omega} + \frac{V_b}{3\Omega} + 4\Omega = 0$$
However when I row reduce them I get $$V_a = -12 \\V_b = -12 $$
but the solution is: enter image description here

Best Answer

Just set it up and knock them down. The left side of the initial equation, for each node I set up, will be the currents leaving the node. The right side will be the currents entering the node.

Node 'a' first:

$$\begin{align*} \frac{V_a}{4\Omega}+\frac{V_a}{2\Omega}+3A&=\frac{0V}{4\Omega}+\frac{V_b}{2\Omega} \\ V_a\cdot\left(\frac{1}{4\Omega}+\frac{1}{2\Omega}\right)+V_b\cdot\left(\frac{-1}{2\Omega}\right)&=-3A \end{align*}$$

Node 'b' now: $$\begin{align*} \frac{V_b}{2\Omega}+\frac{V_b}{3\Omega}&=4A+\frac{V_a}{2\Omega}+\frac{0V}{3\Omega} \\ V_a\cdot\left(\frac{-1}{2\Omega}\right)+V_b\cdot\left(\frac{1}{2\Omega}+\frac{1}{3\Omega}\right)&=4A \end{align*}$$

Those are your two equations in two unknowns. In this case, \$V_a=-1\tfrac{1}{3}V\$ and \$V_b=4V\$, if I did things right.

I can reconstitute my equation form into yours, more or less, as:

$$\begin{align*} \frac{V_a}{4\Omega}+\frac{V_a-V_b}{2\Omega}+3A&=0 \\ \frac{V_b-V_a}{2\Omega}+\frac{V_b}{3\Omega}-4A&=0 \end{align*}$$

Or, in negated form:

$$\begin{align*} -\frac{V_a}{4\Omega}-\frac{V_a-V_b}{2\Omega}-3A&=0 \\ \frac{V_a-V_b}{2\Omega}-\frac{V_b}{3\Omega}+4A&=0 \end{align*}$$

Perhaps you can see the differences here? This last equation pair looks like yours except for the sign of \$\tfrac{V_b}{3\Omega}\$. Probably, because you failed to realize it was \$\tfrac{0V-V_b}{3\Omega}\$, as a guess.

Let me know if you need me to walk slowly through how I set up the initial equations.