Electronic – Nodal analysis on this sort of circuit

impedancesteady state

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.

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They obtain this after mesh analysis.
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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.

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