A 50.0-Ohm resistor, a 0.100-H inductor, and a 10.0-mu F capacitor are connected

Question

A 50.0-Ohm resistor, a 0.100-H inductor, and a 10.0-mu F capacitor are connected in series with an AC generator; the generator produces an RMS voltage of 100V. If the generator is operated at the resonance frequency of the circuit, what will be the resulting average power dissipation?

Explanation / Answer

P = (Irms)^2 R = ((Vrms)^2/Z^2) R >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> resonance ==> Z = R >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ==> P = (Vrms)^2/R = 100*100/50 = 200 W

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