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 P2. The ratio of P1 to P2 will be :

Question

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 P2. The ratio of P1 to P2 will be : (A) 1: 2 (B) 1: 3 (C) 1: 1 (D) 1: √2

Tardigrade Admin · Accepted Answer

P=(R/Z) ⇒ P1=(R/√R2+XL2)=(R/R √2)( text as XL=R) P1=(1/√2) P2=(R/√R2+(XL-XL)2)=P2=1 (P1/P2)=(1/√2)

Q. In a series $L R$ 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 :

$1 : 2$

$1 : 3$

$1 : 1$

$1 : 2$

$P = \frac{R}{Z} \Rightarrow P_{1} = \frac{R}{R ^{2} + X _{L}^{2}} = \frac{R}{R 2} (as X_{L} = R)$
$P_{1} = \frac{1}{2}$
$P_{2} = \frac{R}{R ^{2} + ( X _{L} - X _{L} ) ^{2}} = P_{2} = 1$
$\frac{P _{1}}{P _{2}} = \frac{1}{2}$

Q. In a series LR circuit with XL​=R, power factor is P1​. If a capacitor of capacitance C with XC​=XL​ is added to the circuit the power factor becomes P2​. The ratio of P1​ to P2​ will be :

1:2

1:3

1:1

1:2​

P=ZR​⇒P1​=R2+XL2​​R​=R2​R​( as XL​=R) P1​=2​1​ P2​=R2+(XL​−XL​)2​R​=P2​=1 P2​P1​​=2​1​

Q. In a series $L R$ 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 :

$1 : 2$

$1 : 3$

$1 : 1$

$1 : 2$

$P = \frac{R}{Z} \Rightarrow P_{1} = \frac{R}{R ^{2} + X _{L}^{2}} = \frac{R}{R 2} (as X_{L} = R)$
$P_{1} = \frac{1}{2}$
$P_{2} = \frac{R}{R ^{2} + ( X _{L} - X _{L} ) ^{2}} = P_{2} = 1$
$\frac{P _{1}}{P _{2}} = \frac{1}{2}$