I'm reviewing nodal analysis on sinusoidal steady state circuits which my textbook typically only does mesh analysis on and I can't seem to get to the answer they obtained for an example problem.

They obtain this after mesh analysis.

My thought is that the middle node and top node are my only essential node but I assume that I'm treating it wrong since I'm not obtain the same solution.

Does anyone have any insight here?

This is simply for practice and my own knowledge.

Edit:

I obtained:

$$Io = 1.11897\angle19.5104°$$

The book obtained:

$$Io = 1.465\angle38.48°$$

## Best Answer

I assigned \$V_A\$ to your \$a\$ node. I mentally grounded your \$b\$ node. Given that preface, nodal provides:

$$\begin{align*} \frac{V_A}{5\:\Omega}+\frac{V_A}{20\:\Omega+j 15\:\Omega}&=3\:\textrm{A}+\frac{j40\:\textrm{V}}{5\:\Omega} \end{align*}$$

(There's no need to worry about the "middle node" voltage because it is on the other side of a current source or else you have to go past a voltage source to get there. Either way, it doesn't matter.)

I used \$j40\:\textrm{V}\$ for your voltage source because it's \$90^\circ\$ out of phase and this is achieved by simply multiplying by \$j\$. The current source is \$0^\circ\$ so a simple \$3\:\textrm{A}\$ is fine.

Solve for \$V_A\$:

$$V_A=\frac{3\:\textrm{A}+\frac{j40\:\textrm{V}}{5\:\Omega}}{\frac{1}{5\:\Omega}+\frac{1}{20\:\Omega+j 15\:\Omega}}$$

And then you know that:

$$\begin{align*} I_0&=\frac{V_A}{20\:\Omega+j15\:\Omega}\\\\ &=V_A\cdot\frac{1}{20\:\Omega+j15\:\Omega}\\\\ &=\frac{3\:\textrm{A}+\frac{j40\:\textrm{V}}{5\:\Omega}}{\frac{1}{5\:\Omega}+\frac{1}{20\:\Omega+j 15\:\Omega}}\cdot\frac{1}{20\:\Omega+j15\:\Omega}\\\\ &=\frac{3\:\textrm{A}+\frac{j40\:\textrm{V}}{5\:\Omega}}{\frac{20\:\Omega+j15\:\Omega}{5\:\Omega}+1}\\\\ &=\frac{15\:\textrm{V}+j40\:\textrm{V}}{25\:\Omega+j15\:\Omega}\\\\ &\approx 1.14705882 + j0.911764706\\\\&\approx 1.46528455 \angle 38.4801982^\circ \end{align*}$$

Rounded, the above matches the answer you are supposed to get.