Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACSeries AC CircuitsMedium2 minPYQ_2018
PhysicsMediumsingle choice

An inductor20 mH,capacitor100 μFand a resistor50 Ωare connected in series across a source of emf,V= 10sin⁡(314t⁡). The power loss in the circuit isNEET 2018

Options:

Answer:
C
Solution:

V0=10 V, ω=314  rad/s

P=VrmsIrmscos⁡ϕ

P= V rms V rms Z R Z

= ( V rms ) 2 R Z 2

XL=ωL=314 20×10-3=6.280

XC=1ωC=1314×100×10-6=31.84 Ω

R=50 Ω

Z= ( X C − X L ) 2 + R 2

=  31.84-6.282+502=56 Ω

⇒P= ( 10 2 ) 2 × 50 ( 56 ) 2 =0.79 W

Stream:NEETSubject:PhysicsTopic:EMI/ACSubtopic:Series AC Circuits
⏱ 2mℹ️ Source: PYQ_2018

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