A coil of negligible resistance is connected in series with 90Ω resistor across 120V,60Hz supply. A voltmeter reads 36 V across resistance. Inductance of the coil is :
(A)0.76 H✓ Correct
(B)2.86 H
(C)0.286 H
(D)0.91 H
Answer: (A)
Solution
Current through the resistor: I=RVR=9036=0.4 A (this is the circuit current, coil and resistor in series). Since the coil is purely inductive, V2=VR2+VL2:
VL=1202−362=13104≈114.5 V
XL=IVL=0.4114.5≈286.2Ω
L=2πfXL=2π(60)286.2≈0.76 H
JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Single correct option