Question: While testing a 100KVAR 480V 3 phase wye connected capacitor, what would the expected uF be across A-B terminals?
|a.|1152uF|
|b.|576uF|
|c.|665uF|
|d.|58uF|
I keep coming up with 1152, but the testguy quiz says its 576. Why??
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So you accurately got the equivalent capacitance of 1152 for a single capacitor. Since this is wye connected however, if you measure across A+B, you are not measuring one capacitor, but 2. You’re measuring the cap installed from A to neutral and the Cap installed from neutral to B. The other phase is open so it is two series capacitors across A & B. Capacitors add as if they are parallel resistors when they are in series, so the measured value between terminal A and B is the total capacitance of the series circuit of of cap A and B. Or Ca*Cb/(Ca+Cb) = C^2/2C = C/2 or 1152/2 = 576. If they were delta connected, 1152 would be accurate. Hope that clears it up.
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Your answer is correct (1152). The reason why is:
1- Qtotal = 100KVAR. and each Capacitor in the bank will give you 33.33KVAR.
2- Now Q = V^2/Xc → Xc = V^2/Q → Xc= 277^2 / 33.33KVAR → Xc = 2.3 Ohms Per Capacitor.
3- Xc = 1/2 * Pi * f * C → C ~ 1152 MicroFarad.
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For clarity on this, the answer is 576 due to capacitors in series reciprocate. Since the terminals on A and B are measured and in series, the capacitance of 1152 reciprocates or simply its just half. Same as when two 100 ohm resistors in parallel half to 50 total Resistance.
Seems like everyones math is on point, just one of those small details that NETA likes to trip you with.
I myself was stumped by this one too.
