JEE Main 2014 · 9 Apr · Shift 1 · Physics Q21

← JEE Main 2014 · 9 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 21 of 90

When the rms voltages $V_L$ , $V_C$ and $V_R$ are measured respectively across the inductor L, the capacitor C and the resistor R in a series LCR circuit connected to an AC source, it is found that the ratio $V_L:V_C:V_R=1:2:3$ . If the rms voltage of the AC source is 100 V, then $V_R$ is close to :

50 V

70 V

90 V

100 V

Answer and worked solution

Official answer: C

Let $V_L=k$ , $V_C=2k$ , $V_R=3k$ . Then
$V=\sqrt{V_R^2+(V_L-V_C)^2}=k\sqrt{10}=100\implies k=31.6.$
So $V_R=94.9$ V, closest to 90 V.

Topics:LCR circuits and resonance Reactance and impedance

← Previous question Next question →

JEE Main 2014 · 9 Apr · Shift 1, Physics question 21