# Find power of resistor and source in AC circuit

1. Apr 6, 2013

### Same-same

1. The problem statement, all variables and given/known data
If an average power of 500W is dissipated in the 20Ω resistor, find Vrms, I S RMS, the power factor seen by the source, and the magnitude of VS
(Based on circuit in attached diagram)

2. Relevant equations

Pave= Irms*Vrms*pf*$\frac{1}{2}$
Imaginary number referred to as "j", not "i".

3. The attempt at a solution
VA=V
By a node equation at node A, we see that $\frac{V}{20}$= $\frac{VS}{-j*20}$, so V = VS $\angle$-90

Loop 1: (j*20-j*20)IS -j*20I= VS
By observation, I=V/20, so V=VS$\angle$90.
Loop 2: (20+j*20)I-j*20IS=0, so IS= 1.41$\angle$135 *I
= 1.41$\angle$135 *$\frac{V}{20}$= 0.0705 V $\angle$135

Since I and V are in phase, the power across the resistor is 1
Solve for V:
500=V*$\frac{V}{20}$$\frac{1}{2}$, so V=$\sqrt{20,000}$=
100$\sqrt{2}$=141.4.

Vrms=$\frac{V}{\sqrt{2}}$ =100,

Since V = 100$\sqrt{2}$ IS =7.05 *$\sqrt{2}$ $\angle$135, so I S RMS = 7.05

V=VS$\angle$90, so VS=V$\angle$-90
And IS= 0.0705 V $\angle$135,
So power factor pf = cos(135-90)= 0.707

VS=V$\angle$-90 = -141.4*j, so magnitude given by 141.4.

I think I got this right, but I just want to make sure.

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2. Apr 6, 2013

### Same-same

Just realized the diagram I attached was small, hopefully this one's easier to read.

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3. Apr 6, 2013