Alternating Current - Result Question 13

13. An inductor 20mH, a capacitor 100μF and a resistor 50Ω are connected in series across a source of emf, V=10sin314t. The power loss in the circuit is

(a) 0.79W

(b) 0.43W

(c) 1.13W

(d) 2.74W

[2018]

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Answer:

Correct Answer: 13. (a)

Solution:

  1. (a) Power dissipated in an LCR series circuit connected to an a.c. source of emf E

P=Erms irmscosϕ=Erms2RZ2=Erms2RR2+(ωL1Cω)2

=(102)2×50(50)2+(314×20×1031314×100×106)2

Solving we get, P=0.79W



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