Engineering What is the power factor of an RL circuit with given parameters?

[hogrampage]

Homework Statement

See attachment.

Homework Equations

V(t) = V_mcos(\omegat + \theta)
P(t) = V_mI_mcos(\omegat + \phi)cos\omegat
P_R(t) = \frac{V^{2}(t)}{R}
PF = \frac{P}{V_effI_eff}
S = V_effI_eff
P_avg = I_eff²(Re(Z_eq)

The Attempt at a Solution

(a)
I_eff = 5 A
f = 50 Hz or \omega \approx 314 rad/s
Z_eq = \frac{80(j60)}{80+j60} = 28.8 + j38.4 Ω(48\angle53° Ω)
V = (5\angle0)(48\angle53) = 240\angle53.13010235° V
P_avg = (5)^{2}(28.8) = 720 W

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

(b)
P_avg = 720 W

(c)
P_R(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

P_L(2 ms) = P(2 ms) - P_R(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.

 

[gneill]

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

 

[hogrampage]

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

 

[NascentOxygen]

(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?

 

[hogrampage]

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