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

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Discussion Overview

The discussion revolves around calculating the power factor of an RL circuit given specific parameters and equations. Participants are addressing a homework problem that involves both theoretical calculations and practical applications of electrical engineering concepts.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents calculations for the effective current, impedance, voltage, average power, and power factor, suggesting a power factor of 0.6.
  • Another participant confirms the correctness of the calculations for parts (a) and (b) but questions the interpretation of part (c), indicating a potential misunderstanding of the problem statement.
  • There is a clarification regarding the time variable in part (c), with one participant confirming that it refers to t = 2 ms.
  • Participants express uncertainty about the interpretation of the problem and the specific requirements for part (c).

Areas of Agreement / Disagreement

Participants generally agree on the correctness of parts (a) and (b), but there is disagreement regarding the interpretation of part (c), which remains unresolved.

Contextual Notes

The discussion highlights potential ambiguities in the problem statement, particularly regarding the time variable in part (c), which may affect the calculations and interpretations presented.

Who May Find This Useful

Students and practitioners in electrical engineering, particularly those studying AC circuits and power factor calculations.

hogrampage
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Homework Statement


See attachment.

Homework Equations


V(t) = Vmcos([itex]\omega[/itex]t + [itex]\theta[/itex])
P(t) = VmImcos([itex]\omega[/itex]t + [itex]\phi[/itex])cos[itex]\omega[/itex]t
PR(t) = [itex]\frac{V^{2}(t)}{R}[/itex]
PF = [itex]\frac{P}{V<sub>eff</sub>I<sub>eff</sub>}[/itex]
S = VeffIeff
Pavg = Ieff2(Re(Zeq)

The Attempt at a Solution


(a)
Ieff = 5 A
f = 50 Hz or [itex]\omega[/itex] [itex]\approx[/itex] 314 rad/s
Zeq = [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
Pavg = (5)[itex]^{2}[/itex](28.8) = 720 W

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

(b)
Pavg = 720 W

(c)
PR(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

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.
 

Attachments

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The image is not appearing for me. Can you try to post it again?
 
It should show up now (attached to the first post).
 
(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?
 
Yeah, it is t = 2 ms. For some reason, it cut off half of it.
 

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