Electronics - RLC Circuits and Resonance - Discussion

Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 41)
41.
What is the power factor?

0.279
0.479
0.679
0.879
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
14 comments Page 1 of 2.

Murugan said:   7 years ago
In parallel rlc circuit.

Z=1/√((1÷r)^2+(1÷xl-1÷xc)^2).
i=v/z.
Power factor=true power/apparent power.
True power=i^2*r.
Apparent power=Vs * i.
Therefore power factor=i^2*r/Vs*i.

@Gasim your formula is a series circuit.

Vaishnavi said:   1 decade ago
Iz = sqrt (Ir^2+Ix^2) = 0.1411 A.

Z = E/Iz = 497/0.1411 = 3522.3 ohm.

True power = Ir^2*R = 61.702 W.

App power = Iz^2*Z = 70.126 W.

Power factor = 61.702/70.126

= 0.879.

Kalvs said:   2 years ago
For true power P, I used V^2/R /= 497^2/4k = 61.75 W.
Apparent Power, I used V^2/Z = 497^2/3516 = 70.23 VA.
Pf = P/S (real power/apparent power) = 61.75/70.23 = 0.879.
(1)

Minu vaishnavi said:   1 decade ago
For parallel RLC circuit powerfactor=(truepower)/(apparentpower)
True power (Pr)=Ir^2*R
Apparent power (Pa)=Vs*It
P.F=(Ir^2*R)/(Vs*It)

Shawn said:   1 decade ago
In previous question, we have calculated phase angle -28.5.
And Power Factor = cos(@).
Therefore ..cos (28.5) = 0.879.
Option D?

Abhiee said:   10 years ago
I'm also getting 0.879. What is the correct answer and solution for this problem? please mention.

Vandana said:   1 decade ago
@Neeraja

please tell me,how the answer can be evaluated using this formula.

Alvin said:   7 years ago
I Can't get this. Please, someone, explain in clear.

Gasim said:   1 decade ago
z=sqrt(R^2+(xl-xc)^2)
i=vs/z
p=i^2*R
P.F=P/(vs*i)

Pritam said:   5 years ago
PF= R/Z (for series).
pF= Z/R (for Parallel).


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