circuit analysispower
Image source – Fundamentals of Electric Circuit by Alexander & Sadiku, Practice problem 11.15.
I tried to solve the math in the following way.
\begin{equation}
S_{old}=140000\angle cos^{-1}(0.85) = 119000+j73749.576\\
S_{new}=140000\angle cos^{-1}(1.00) = 140000+j0.00 \;\;\;\;\;\;\;\;\;\; \\
So, Q_c = 73749.576 \\
And,\; C=\frac{Q_c}{2\pi f {V_{RMS}} ^2} = \frac{73749.576}{2\pi 60 (110) ^2} = 0.0161675\;F
\end{equation}
Which is a wrong answer. Can anyone provide me the correct way?
The node voltage equations are not so difficult. But it's really a tedious work to type the equations. I got the same result with you, maybe we are both wrong :).
Now, part (b) and (c) seem easy enough to answer given enough information but I cannot figure out whether I'm understanding (a) right.
The question reads to me as if part(a) applies before the correction so....
Try looking up power factor phasor diagrams: -
You have an apparent power (volt*amps or VA) of 150 and, you have an angle of 24 degrees. Just apply pythagoras to calculate real (active) power. As bonus you can even calculate reactive power.
Best Answer
They tell you it is 140 kVAR, so you need to determine capacitance that will provide 140 kVAR.
XC = 110^2/140,000
From that you can easily calculate C.