5
\$\begingroup\$

enter image description here
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?

\$\endgroup\$
6
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$
  • 2
    \$\begingroup\$ Opps. I didn’t notice that the unit is in VAR. I did the math thinking its KVA. Thanks. \$\endgroup\$ – Sadat Rafi Jun 12 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\$ – a concerned citizen Jun 12 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\$ – relayman357 Jun 12 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.