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Consider the following R-L-C circuit in which R = 12 Omega, X_(L) =24Omega, X_(C) =8Omega The emf of source is given by V =10 sin (100pit)V (a) Find the energey dissipated in 10 min (b) If resistance is removed from the circuit and value of inductance is doubled express variation of current with time t in the new circuit . . |
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Answer» Solution :Inpedance `Z = sqrt(R^(2) + (X_(L) - X_(C)^(2))) = sqrt((12)^(2) + (24 - 8)^(2)) = 20Omega` `cops phi = ( R)/(Z) = (12)/(20) = (3)/(5)` Power consumed ` P = (1)/(2) V_(0)i_(0) cos phi = (1)/(2) V_(0) . V_(0)/(Z) cos phi` ` = (1)/(2) (10)^(2)/((20)) xx (3)/(5) = 1.5W` Energy dissipated `E = Pt = (1.5)(10 xx 60) =900J` `Z = X_(L) - X_(C) = 48 -8 =40 Omega` `i_(0) = (V_(0))/(Z) = (10)/(40) = (1)/(4) A` Current lags the voltage by `pi//2` `i = i_(0) SIN (omegat -pi//2)` `= (1)/(4) sin (100 pit - 2)`   .
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