Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 41)
41.
What is the power factor?
Discussion:
14 comments Page 1 of 2.
Arun sundar said:
6 years ago
Use PF=Z/R for the parallel circuit.
(1)
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.
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)
Neeraja Pamala said:
1 decade ago
Power Factor = True Power/Apparent Power
Vandana said:
1 decade ago
@Neeraja
please tell me,how the answer can be evaluated using this formula.
please tell me,how the answer can be evaluated using this formula.
Deep said:
1 decade ago
pf=R/Z
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)
True power (Pr)=Ir^2*R
Apparent power (Pa)=Vs*It
P.F=(Ir^2*R)/(Vs*It)
Gasim said:
1 decade ago
z=sqrt(R^2+(xl-xc)^2)
i=vs/z
p=i^2*R
P.F=P/(vs*i)
i=vs/z
p=i^2*R
P.F=P/(vs*i)
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?
And Power Factor = cos(@).
Therefore ..cos (28.5) = 0.879.
Option D?
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.
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.
Abhiee said:
10 years ago
I'm also getting 0.879. What is the correct answer and solution for this problem? please mention.
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