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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