JEE Main 2022 · 26 Jul · Shift 1 · Physics Q37

← JEE Main 2022 · 26 Jul · Shift 1

PhysicsSingle correct+4 / -1Question 37 of 90

In a series LR circuit $X_L=R$ and power factor of the circuit is $P_1$ . When capacitor with capacitance C such that $X_L=X_C$ is put in series, the power factor becomes $P_2$ . The ratio $\dfrac{P_1}{P_2}$ is :

\dfrac12

\dfrac1{\sqrt2}

\dfrac{\sqrt3}{\sqrt2}

2 : 1

Answer and worked solution

Official answer: B

Power factors. $P_1=\frac R{\sqrt2R}=\frac1{\sqrt2}$ ; at resonance $P_2=1$ .

(B)

Topics:Power in AC circuits

← Previous question Next question →

JEE Main 2022 · 26 Jul · Shift 1, Physics question 37