Question

If an inductor with inductive reactance, XL=R is connected in series with resistor R across an A.C voltage, power factor comes out to be P1. Now, if a capacitor with capacitive reactance, XC=R is also connected in series with inductor and resistor in the same circuit, power factor becomes P2. Find $\frac{P_1}{P_2}$

• A. $\sqrt2:1$
• B. $1:\sqrt2$
• C. 1 : 1
• D. 1 : 2

Answer 1

Answer: (B)

$1:\sqrt2$

Answer 2

$Z=\sqrt{R^2+R^2}$
$=\sqrt{2}R$
$P_1=\cos\phi=\text{power factor}=\frac{R}{Z}=(\frac{1}{\sqrt2})$
When capacitor is also connected in series
The LCR circuit is in resonance stage
So, $Z=\sqrt{R^2+(X_L-X_C)^2}$
Z = R
$P_2=\cos\phi=\text{power factor}=\frac{R}{Z}=\frac{R}{R}=1$
So , $\frac{P_1}{P_2}=\frac{(\frac{1}{\sqrt2})}{1}=\frac{1}{\sqrt2}$

So, the correct answer is (B) : $1:\sqrt2$