Question 385: AC fundamentals

Question

In a series LR circuit, $$X_L=R$$, 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 $$P_1$$ to $$P_2$$ will be

Accepted Answer

$$\begin{array}{ll} 
& 	ext { Power factor }=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2}} \
& 	ext { Given: } \mathrm{X}_{\mathrm{L}}=\mathrm{R} \
	herefore \quad & \mathrm{P}_1=\frac{\mathrm{R}}{\sqrt{2 \mathrm{R}^2}}=\frac{1}{\sqrt{2}}
\end{array}$$
When $$\mathrm{C}$$ is connected, the circuit becomes series LCR circuit.
$$	herefore \quad 	ext { Power factor }=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}ight)^2}}$$
Given : $$\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}$$
$$\begin{array}{ll}
	herefore & \mathrm{P}_2=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2}}=1 \
	herefore & \mathrm{P}_1: \mathrm{P}_2=1: \sqrt{2}
\end{array}$$

Answer

$$1: 3$$

Answer

$$1: \sqrt{2}$$

Answer

$$1: 1$$

Answer

$$1: 2$$

AC fundamentals

In a series LR circuit, $X_L=R, power factor is P_1. If a capacitor of capacitance C with X_C=X_L is added to the circuit the power factor becomes P_2. The ratio of P_1 to P_2$ will be

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