# Power Factor of RL circuit

1. Oct 1, 2012

### hogrampage

1. The problem statement, all variables and given/known data
See attachment.

2. Relevant equations
V(t) = Vmcos($\omega$t + $\theta$)
P(t) = VmImcos($\omega$t + $\phi$)cos$\omega$t
PR(t) = $\frac{V^{2}(t)}{R}$
PF = $\frac{P}{VeffIeff}$
S = VeffIeff
Pavg = Ieff2(Re(Zeq)

3. The attempt at a solution
(a)
Ieff = 5 A
f = 50 Hz or $\omega$ $\approx$ 314 rad/s
Zeq = $\frac{80(j60)}{80+j60}$ = 28.8 + j38.4 Ω(48$\angle$53° Ω)
V = (5$\angle$0)(48$\angle$53) = 240$\angle$53.13010235° V
Pavg = (5)$^{2}$(28.8) = 720 W

PF = $\frac{720}{1200}$ = 0.6

(b)
Pavg = 720 W

(c)
PR(2 ms) = $\frac{57600cos^{2}(314(0.002)+53°)}{80}$ $\approx$ 253 W
-tan($\frac{{\omega}L}{R}$) $\approx$ -36.85°
P(2 ms) = 1200cos(314(0.002)-36.85°)cos(314(0.002)) $\approx$ 968 W

PL(2 ms) = P(2 ms) - PR(2 ms) = 715 W

(d)
Have not done yet.

I want to know if I am going in the right direction or if I'm doing it all completely wrong haha. I don't want to do part (d) until I know the other parts are headed in the right direction.

Any help is appreciated.

#### Attached Files:

• ###### imgPF.jpg
File size:
40.7 KB
Views:
67
Last edited: Oct 1, 2012
2. Oct 1, 2012

### Staff: Mentor

The image is not appearing for me. Can you try to post it again?

3. Oct 1, 2012

### hogrampage

It should show up now (attached to the first post).

4. Oct 2, 2012

### Staff: Mentor

(a) and (b) are correct. I haven't figured out what is being asked in (c). I read it as being asked to express instantaneous power as a function of time, specifically t - 2msec, but you seem to have read it as t = 2msec. Is precisely half of the equals sign missing?

5. Oct 2, 2012

### hogrampage

Yeah, it is t = 2 ms. For some reason, it cut off half of it.