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

[hogrampage]

Homework Statement

See attachment.

Homework Equations

V(t) = V_mcos([itex]\omega[/itex]t + [itex]\theta[/itex])
P(t) = V_mI_mcos([itex]\omega[/itex]t + [itex]\phi[/itex])cos[itex]\omega[/itex]t
P_R(t) = [itex]\frac{V^{2}(t)}{R}[/itex]
PF = [itex]\frac{P}{V_effI_eff}[/itex]
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 [itex]\omega[/itex] [itex]\approx[/itex] 314 rad/s
Z_eq = [itex]\frac{80(j60)}{80+j60}[/itex] = 28.8 + j38.4 Ω(48[itex]\angle[/itex]53° Ω)
V = (5[itex]\angle[/itex]0)(48[itex]\angle[/itex]53) = 240[itex]\angle[/itex]53.13010235° V
P_avg = (5)[itex]^{2}[/itex](28.8) = 720 W

PF = [itex]\frac{720}{1200}[/itex] = 0.6

(b)
P_avg = 720 W

(c)
P_R(2 ms) = [itex]\frac{57600cos^{2}(314(0.002)+53°)}{80}[/itex] [itex]\approx[/itex] 253 W
-tan([itex]\frac{{\omega}L}{R}[/itex]) [itex]\approx[/itex] -36.85°
P(2 ms) = 1200cos(314(0.002)-36.85°)cos(314(0.002)) [itex]\approx[/itex] 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.