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Question: Answered & Verified by Expert
An inductor 20 mH, capacitor 100 μF and a resistor 50 Ω are connected in series across a source of emf, V= 10sin⁡(314t⁡). The power loss in the circuit is
PhysicsAlternating CurrentNEETNEET 2018
Options:
  • A 2.74W
  • B 0.43W
  • C 0.79W
  • D 1.13W
Solution:
2853 Upvotes Verified Answer
The correct answer is: 0.79W
V0=10 V, ω=314  rad/s

P=VrmsIrmscosϕ

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

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