InterviewSolution
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InterviewSolution
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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 isA. 960 JB. 900 JC. 250JD. 500J |
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Answer» Correct Answer - 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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