- #1
Same-same
- 27
- 0
Homework Statement
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)
Homework Equations
Pave= Irms*Vrms*pf*[itex]\frac{1}{2}[/itex]
Imaginary number referred to as "j", not "i".
The Attempt at a Solution
VA=V
By a node equation at node A, we see that [itex]\frac{V}{20}[/itex]= [itex]\frac{VS}{-j*20}[/itex], so V = VS [itex]\angle[/itex]-90
Loop 1: (j*20-j*20)IS -j*20I= VS
By observation, I=V/20, so V=VS[itex]\angle[/itex]90.
Loop 2: (20+j*20)I-j*20IS=0, so IS= 1.41[itex]\angle[/itex]135 *I
= 1.41[itex]\angle[/itex]135 *[itex]\frac{V}{20}[/itex]= 0.0705 V [itex]\angle[/itex]135
Since I and V are in phase, the power across the resistor is 1
Solve for V:
500=V*[itex]\frac{V}{20}[/itex][itex]\frac{1}{2}[/itex], so V=[itex]\sqrt{20,000}[/itex]=
100[itex]\sqrt{2}[/itex]=141.4.
Vrms=[itex]\frac{V}{\sqrt{2}}[/itex] =100,
Since V = 100[itex]\sqrt{2}[/itex] IS =7.05 *[itex]\sqrt{2}[/itex] [itex]\angle[/itex]135, so I S RMS = 7.05
V=VS[itex]\angle[/itex]90, so VS=V[itex]\angle[/itex]-90
And IS= 0.0705 V [itex]\angle[/itex]135,
So power factor pf = cos(135-90)= 0.707
VS=V[itex]\angle[/itex]-90 = -141.4*j, so magnitude given by 141.4.
I think I got this right, but I just want to make sure.