An inductor 20 mH, capacitor 100 μF and a resistor 50 Ω are connected in series across a source of emf, V =…

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
  1. 2.74W
  2. 0.43W
  3. 0.79W
  4. 1.13W

Solution

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

Asked in: NEET 2018

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