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In an ac circuit the instantaneous values of current and applied voltage are respectively `i=2(Amp) sin (250pis^(-1)t and epsi =(10V) sin [(250pis^(-1))t+(pi)/(3)`, Find the instantaneous power drawn from the source at `t=(2)/(3)ms` and its average value. |
Answer» Correct Answer - `10 W, 5W` `P=iepsilon=[2 sin250pit][10sin(250pit+pi//3)]` at `t=2/3xx10^(-3)rArr P=10 watt` `ltPgt = i_(rms)epsilon_(rms)cos phi=2/sqrt2xx10sqrt2xxcos pi//3` `P=iepsilon=[2sin 250pi t][10 sin (250pit+pi//3)] , t=2/3 xx10^(-3) rArr P=10` watt `ltPgt = i_(rms)epsilon_(rms)cos phi=2/sqrt2xx10sqrt2xxcos pi//3` |
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