# Determine the absorbed complex power (AC Circuit Power Analysis)

I've been trying to solve a book problem from Hayt, et al. Engineering Circuit Analysis (8th Ed.) —Practice Problem 11.9.

And here's my latest attempt, I just can't see what I'm doing wrong, why I don't get the answers from the book. Probably it's just some blunder I don't realize I'm making like a sign or a rule poorly applied. Please help me understand what I'm doing wrong.

$$\V_1\$$ is the voltage of the 1 $$\\Omega\$$ resistor.

• To point you in the right direction, I would combine impedances to find I1 first. This should give you 5.16A∠25.46 which gives you the correct value for the power consumed by the 1 Ohm resistor.
– vir
Commented Jul 26, 2022 at 23:07
• @vir thank you, but I have a question with respect to my nodal approach, I think I'm doing wrong the relation for the voltage of the 1 Ohm resistor, right? Commented Jul 26, 2022 at 23:17

The input impedance of the circuit is given by:

$$\underline{\text{Z}}_{\space\text{i}}=1+\left(\left(5+10\text{j}\right)\space\text{||}\space-10\text{j}\right)=21-10\text{j}\tag1$$

Where $$\\alpha\space\text{||}\space\beta:=\frac{\alpha\beta}{\alpha+\beta}\$$.

So, the input current is given by:

$$\underline{\text{I}}_{\space\text{i}}=\frac{\underline{\text{V}}_{\space\text{i}}}{\underline{\text{Z}}_{\space\text{i}}}=\frac{120\exp\left(0\text{j}\right)}{21-10\text{j}}=\frac{2520}{541}+\frac{1200\text{j}}{541}\tag2$$

So, we get:

$$\text{S}_{1\space\Omega}=\left|\frac{2520}{541}+\frac{1200\text{j}}{541}\right|\cdot\left|1\cdot\left(\frac{2520}{541}+\frac{1200\text{j}}{541}\right)\right|=\frac{14400}{541}\approx26.6174\space\text{VA}\tag3$$