Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACSeries AC CircuitsEasy2 minPYQ_2020
PhysicsEasysingle choice

A100 V ACsource of frequency500 Hzis connected to aLCRcircuit withL=8.1 mH,C=12.5 μFandR=10 Ω, all connected in series. The potential difference across the resistance is

Options:

Answer:
A
Solution:

Z=R2+XL-XC2

HereXL=2πfL=2×3.14×500×8.1×10-3=25.4 Ω

andXC=12πfC=12×3.14×500×12.5×10-6=25.4 Ω

∴  Z=102+25.4-25.42=10 Ω

Nowirms=ErmsZ=10010=10 A

∴    VR=irms×R=10×10=100 V

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:EMI/ACSubtopic:Series AC Circuits
⏱ 2mℹ️ Source: PYQ_2020

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