Alternating Current Question 10
Question 10 - 30 January - Shift 1
In a series LR circuit with $X_L=R$. 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 :
(1) $1: 3$
(2) $1: \sqrt{2}$
(3) $1: 1$
(4) $1: 2$
Show Answer
Answer: (2)
Solution:
$P=\frac{R}{Z} \Rightarrow P_1=\frac{R}{\sqrt{R^{2}+X_L^{2}}}=\frac{R}{R \sqrt{2}}(.$ as $.X_L=R)$
$P_1=\frac{1}{\sqrt{2}}$
$P_2=\frac{R}{\sqrt{R^{2}+(X_L-X_L)^{2}}}=P_2=1$
$\underline{P_1}=\underline{1}$
