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In a series LR circuit with \mathrm{X}_{\mathrm{L}}=\mathrm{R},power factor is P_{1}. If a capacitor of capacitance C with x_C=x_L is added to the circuit the power factor becomes P_{2}. The ratio of P_{1} toP_{2} will be :

Option: 1

1:3


Option: 2

1: 2


Option: 3

1:\sqrt{2}


Option: 4

1: 1


Answers (1)

best_answer

\text { Power factor }=\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}

\mathrm{P}_1=\frac{\mathrm{R}}{\sqrt{\mathrm{x}_{\mathrm{L}}^2+\mathrm{R}^2}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{R}^2}}=\frac{\mathrm{R}}{\sqrt{2} \mathrm{R}}=\frac{1}{\sqrt{2}}

\begin{aligned} & P_2=\frac{R}{Z}=\frac{R}{\sqrt{\left(x_L-x_C\right)^2+R^2}}=\frac{R}{R}=1 \\ & \frac{P_1}{P_2}=\frac{1}{\sqrt{2}} \end{aligned}

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Gaurav

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