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An inductor 20xx10^(-3) a capacitor 100(...

An inductor `20xx10^(-3)` a capacitor `100(mu)F` and a resistor `50 Omega` are connected in series across a source of emf `V=10 sin 314 t`.
Then the energy dissipated in the circuit in 20 mim is

A

960 J

B

900 J

C

250J

D

500J

Text Solution

Verified by Experts

The correct Answer is:
A
`Z^(2)=(X_(C )-X_(L))^(2) + R^(2)=(31.85 - 6.28)^(2) + (50)^(2) = 3154`
`P=((E_(rms)^(2))/(Z^(2))) R= ((10//sqrt(2))^(2))/(3154) xx 50 = 0.8 W`
Heat produced in 20 min `=(0.8)(20xx60)=960 J`
`X_(C )-X_(L) = 31.85 - 2(6.28)=19.29`
`I_(m)=(10)/(19.29)=0.52`
Hence, `I=0.52 sin (314t+(pi)//(2))=0.52 cos 314 t`.
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Knowledge Check

  • An inductor 20 xx 10^(-3) henry, a capacitor 100 mu F and a resistor 50 Omega are connected in series across a source of emf V = 10 sin 314 t . (a) The energy dissipated in the circuit in 20 minutues is 951 J . (b) If resistance is removed from the circuit and the value of inductance is doubled, then the varation of current with time in the new circuit is 0.52 cos (314 t)

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    C
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    D
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