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  • 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 equ

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:

Option: 1

31 \mu \mathrm{F}


Option: 2

54 \mu \mathrm{F}


Option: 3

151 \mu \mathrm{F}


Option: 4

201 \mu \mathrm{F}


Answers (1)

best_answer

Resistance of circuit, \mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{110 \times 110}{330}=\frac{110}{3} \Omega
\mathrm{1^{\text {st }} case} Power factor, \mathrm{\cos \phi=0.6}
Since, current lags the voltage thus, the circuit contains resistance and inductance.
\mathrm{\begin{aligned} & \therefore \cos \phi=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2}}=0.6 \\ & \Rightarrow \mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2=\left(\frac{\mathrm{R}}{0.6}\right)^2 \\ & \Rightarrow \mathrm{X}_{\mathrm{L}}^2=\frac{\mathrm{R}^2}{(0.6)^2}-\mathrm{R}^2 \\ & \Rightarrow \mathrm{X}_{\mathrm{L}}^2=\frac{\mathrm{R}^2 \times 0.64}{0.36} \\ & \therefore \quad \mathrm{X}_{\mathrm{L}}=\frac{0.8 \mathrm{R}}{0.6}=\frac{4 \mathrm{R}}{3} \end{aligned}}

\mathrm{2^{\text {nd }} }case Now \mathrm{ \cos \phi=1 }
Given
Therefore, circuit is purely resistive, i.e., it contains only resistance. This is the condition of resonance in which  
\mathrm{ \begin{aligned} & \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \\ \therefore & \mathrm{X}_{\mathrm{C}}=\frac{4 \mathrm{R}}{3}=\frac{4}{3} \times \frac{110}{3}=\frac{440}{9} \Omega \\ \Rightarrow & \frac{1}{2 \pi \mathrm{fC}}=\frac{440}{9} \Omega \\ \therefore & \mathrm{C}=\frac{9}{2 \times 3.14 \times 60 \times 440}=0.000054 \mathrm{~F}=54 \mu \mathrm{F} \end{aligned} [from eq. (i)] }

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Nehul

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