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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?


1 Answer 1


They tell you it is 140 kVAR, so you need to determine capacitance that will provide 140 kVAR.

$$X_C = \frac{110^2}{140,000}$$

From that you can easily calculate C.

  • 2
    \$\begingroup\$ Opps. I didn’t notice that the unit is in VAR. I did the math thinking its KVA. Thanks. \$\endgroup\$
    – Sadat Rafi
    Commented Jun 12, 2020 at 21:57
  • \$\begingroup\$ You were right, but I still find the text to be misleading: "Find the value [...] needed to correct a load of [...]" tells me that the load is 140 kVAr, and that needs to be corrected. Otherwise, I would have said "Find the value [...] that is used for a compensated power of [...]". It seems to convey a clearer message, but maybe it's just me (most possible). \$\endgroup\$ Commented Jun 12, 2020 at 22:06
  • \$\begingroup\$ Yeah, i know what you’re saying. But, the load is fully specified by saying any of the following: X kVAR-0.85pf lag, X kW-0.85pf lag, or X kVA-0.85pf lag. \$\endgroup\$ Commented Jun 12, 2020 at 22:14

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