What is the power factor in a series circuit?

AI Thread Summary
The discussion centers on calculating the power factor in a series circuit, with one participant confirming their answer while questioning the method used. The calculation involves the relationship between real power (PR), average power (Pavg), and impedance (Z). There is some confusion regarding the precision of the value 33.36 versus 33.33 in the calculations. Ultimately, the correct formula for the power factor is established as cos(φ) = R/Z. The conversation concludes with an acknowledgment of understanding the concept.
hidemi
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Homework Statement
A series circuit consists of a 15-Ω resistor, a 25-mH inductor, and a 35-μF capacitor. If the frequency is 100 Hz the power factor is:

The answer is 0.45
Relevant Equations
PR / Pavg = Irms^2 * R / Irms^2*Z
I calculated in the following and got the correct answer. However, I wonder whether this way is correct or not. Thanks!

PR / Pavg = Irms^2 * R / Irms^2*Z = 15 /33.36 = 0.45
 
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Hi,

hidemi said:
and got the correct answer
So how did you get the 33.36 ? From 15/0.45 ? Then the answer is per se correct, but 'the way' is not :wink:

And you have me wondering why you type 33.36 and not 33.33 :cool:

##\ ##
 
BvU said:
Hi,

So how did you get the 33.36 ? From 15/0.45 ? Then the answer is per se correct, but 'the way' is not :wink:(15^2

And you have me wondering why you type 33.36 and not 33.33 :cool:

##\ ##
z=√[15^2 + (100*2π*25*10^-3 - 1/(100*2π*35*10^-6))^2]
BvU said:
Hi,

So how did you get the 33.36 ? From 15/0.45 ? Then the answer is per se correct, but 'the way' is not :wink:

And you have me wondering why you type 33.36 and not 33.33 :cool:

##\ ##
My detailed calculation is as attached.
 

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Yes the power factor is ##\cos\phi=\frac{R}{Z}##.
 
Delta2 said:
Yes the power factor is ##\cos\phi=\frac{R}{Z}##.
Ok Thank you, I understand.
 
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