JEE Main 2022 · 29 Jul · Shift 2 · Physics Q38

← JEE Main 2022 · 29 Jul · Shift 2

PhysicsSingle correct+4 / -1Question 38 of 90

A circuit element X when connected to an a.c. supply of peak voltage 100 V gives a peak current of 5 A which is in phase with the voltage. A second element Y when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by $\dfrac\pi2$ . If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

\dfrac{10}{\sqrt2}

\dfrac5{\sqrt2}

5\sqrt2

\dfrac52

Answer and worked solution

Official answer: D

Elements. X is a 20 Ω resistor and Y a pure inductor with $X_L=20$ Ω. In series $Z=20\sqrt2$ Ω, so $I_0=\frac{100}{20\sqrt2}=\frac5{\sqrt2}$ A and $I_{rms}=\frac52$ A.

(D)

Topics:Reactance and impedance LCR circuits and resonance

← Previous question Next question →

JEE Main 2022 · 29 Jul · Shift 2, Physics question 38