Question 90: AC fundamentals

Question

In a series LR circuit with $$\mathrm{X_L=R}$$, power factor P
1
. If a capacitor of capacitance C with $$\mathrm{X_C=X_L}$$ is added to the circuit the power factor becomes P
2
. The ratio of P
1
to P
2
will be :

Accepted Answer

$${X_L} = R$$
$$ \Rightarrow {P_1} = {R \over {\sqrt {X_L^2 + {R^2}} }} = {1 \over {\sqrt 2 }}$$
Now, $${X_L} = {X_C} = R$$
$$ \Rightarrow {P_2} = {R \over {\sqrt {{R^2} + {{({X_L} - {X_C})}^2}} }} = 1$$
$$ \Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}$$

Answer

1 : $$\sqrt2$$

Answer

1 : 3

Answer

1 : 2

Answer

1 : 1

AC fundamentals

In a series LR circuit with $\mathrm{X_L=R}, power factor P 1 . If a capacitor of capacitance C with \mathrm{X_C=X_L}$ is added to the circuit the power factor becomes P 2 . The ratio of P 1 to P 2 will be :

Choose the correct answer: