A circuit draws 330 W from a 110 V , 60 Hz AC line. The power factor is 0.6 and the current lags the voltage. The capacitance of a series capacitor th

Question

A circuit draws 330 W from a 110 V , 60 Hz AC line. The power factor is 0.6 and the current lags the voltage. The capacitance of a series capacitor th

A circuit draws 330 W from a 110 V , 60 Hz AC line. The power factor is 0.6 and the current lags the voltage. The capacitance of a series capacitor that will result in a power factor of unity is equal to
A. 31 `mu`F
B. 54 `mu`F
C. 151 `mu`F
D. 201 `mu`F

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - B
Resistance of circuit,
R = `V^(2)/P = 110 xx 110 / 330 = 110/3Omega`
1st case Power factor, cos`phi` = 0.6
Since, current lags the voltage thus, the circuit contains resistance and inductance .
`therefore` `cosphi = R / sqrt (R^(2) + X_L^(2))= 0.6`
`implies R^(2) + X_L^(2) = (R/0.6)^(2)`
`impliesX_L^(2) = R^(2)/(0.6)^(2) - R^(2)`
`implies X_L^(2) = R^(2) xx 0.64/0.36`
`therefore X_L = 0.8R/0.6 = 4R/3` ...... (i)
IInd case Now `cosphi = 1` [Given]
Therefore, circuit is purely resistive, i.e., it contains only resistance. This is the condition of resonance in which
`X_L = X_C`
`therefore X_C = 4R/3 = 4/3 xx 110/3 = 440/9Omega` [from Eq.(i)]
`implies 1/2pifC = 440/9 Omega`
`therefore C = 9 / 2 xx 3.14 xx 60 xx 440` = 0.000054F = 54`mu`F