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Electrostatic Potential And Capacitance

Question
CBSEENPH12039811
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An inductor of 20 mH, a capacitor of 50 μF and a resistor of 40 ohm are connected in series across a source of emf V = 10 sin 340 t. the power loss in the ac circuit is,

  • 0.67 W

  • 0.76 W

  • 0.89 W

  • 0.51 W

Solution

D.

0.51 W

Given,
Inductance, L = 20 mH
Capacitance, C = 50 μF
Resistance, R = 40 ohm
emf, V = 10 sin 340 t
Power loss in AC circuit will be given as,
Pav = IV2 R = open square brackets straight E subscript straight V over straight Z close square brackets squared. R
= open parentheses fraction numerator 10 over denominator square root of 2 end fraction close parentheses squared.40 open square brackets fraction numerator 1 over denominator 40 squared plus open parentheses 340 space x space 20 x 10 cubed space minus begin display style bevelled fraction numerator 1 over denominator 340 x 50 x 10 to the power of negative 6 end exponent end fraction end style close parentheses squared end fraction close square brackets
equals space 100 over 2 x 40 x fraction numerator 1 over denominator 1600 plus left parenthesis 6.8 minus 58.8 right parenthesis squared end fraction equals fraction numerator 2000 over denominator 1600 space plus space 2704 end fraction almost equal to space 0.46 space W almost equal to 0.51 space W

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