JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 19)
In a series LR circuit with $$\mathrm{X_L=R}$$, power factor P1. If a capacitor of capacitance C with $$\mathrm{X_C=X_L}$$ is added to the circuit the power factor becomes P2. The ratio of P1 to P2 will be :
1 : $$\sqrt2$$
1 : 3
1 : 2
1 : 1
Explanation
$${X_L} = R$$
$$ \Rightarrow {P_1} = {R \over {\sqrt {X_L^2 + {R^2}} }} = {1 \over {\sqrt 2 }}$$
Now, $${X_L} = {X_C} = R$$
$$ \Rightarrow {P_2} = {R \over {\sqrt {{R^2} + {{({X_L} - {X_C})}^2}} }} = 1$$
$$ \Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}$$
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