Increasing Power factor via series capicitance

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The discussion revolves around increasing the power factor of a 1.60×10^4 V rms, 60 Hz electrical substation currently delivering 7.00 MW at a power factor of 0.920. The user successfully calculated the rms current but struggles with determining the required series capacitance to achieve a power factor of 1. They note that at a power factor of 1, the reactance of the inductor should equal that of the capacitor, indicating resonance. However, the user lacks the inductance value needed to calculate the capacitance. A suggestion is made to simplify the problem by focusing on the impedance seen by the substation.
hsainz
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Homework Statement



problem: You're the operator of a 1.60×104 V rms, 60 Hz electrical substation. When you get to work one day, you see that the station is delivering 7.00 MW of power with a power factor of 0.920.
a) what is the rms current leaving the station?
b)how much series capacitance should you add to increase the power factor to 1 (give answer in microfarads)
c)how much power will the station then be delivering.

Homework Equations


P=IVcos(θ)(both rms)
tan(θ)=(Xl-Xc)/R
resonance frequency=1/[(LC)^.5]

The Attempt at a Solution


I got the first part of the question. But I am having trouble trying to solve part b. I believe when the power factor is equal to 1, the reactance of the inductor should be equal to that of the capacitor because they will be at the resonance frequency and the phase between the current and the voltage in he circuit will be 0 thus giving a power factor of one. However, i don't know how to find the capacitance required to reach the resonance frequency because i don't have the value of the inductance (L). All i know is that all the ciruit elements share the same current which is I=Vsource/(R^2+(Xl-Xc)^2)^1/2 but I am not sure how to go about solving for the capictance. I've been on theis problem for about 2 hours...can someone please help lol -___-
 
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You're overcomplicating things. What is the impedance (real and imaginary) which the substation sees?
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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