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InterviewSolution
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An `L-C-R` series circuit with `100Omega` resistance is connected to an `AC` source of `200V` and angular frequency `300rad//s`. When only the capacitance is removed, the current lags behind the voltage by `60^@`. When only the inductance is removed the current leads the voltage by `60^@`. Calculate the current and the power dissipated in the `L-C-R` circuitA. `200 W`B. `100 W`C. `50 W`D. `400 W` |
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Answer» Correct Answer - D Av.Electric field energy = `((1)/(2) CV_("rms")^(2)) = 25 xx 10^(-3) J` `:. (1)/(2) C (I_("rms") X_(C ))^(2) = 25 xx 10^(-3) J` `:. (1)/(2) C I_(0)^(2) (1)/(2 pi^(2) v^(2) c^(2)) = 25 xx 10^(-3) J` `:. C = 20 mu F` Average magnetic energy `((1)/(2) LI_("rms")^(2)) = L = 2 H` `:. L = (2 xx 5 xx 10^(-3))/((.10)^(2)) L = 1` henry `V_(R ) = I_("rms") R = V_(C ) = 50 V = V_(L) = I_("rms") L` `= (0.10) .300 = (.10) xx (1)/(2 pi ((50)/(sqrt(t))).20 xx 10^(-6))` `= (.10) xx 2 pi (50)/(sqrt(2)) (I)` `V_(R ) = 30 V = V_(C ) = 50 V = V_(L) = 10 W` rsm voltageof source `E_("rms") = (50)/(sqrt(2))` `:. E_("rms") = 35.36 V` |
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