Saved Bookmarks
| 1. |
(a) For a given a.c. I = I_(m) sin omega tshow that the average power dissipated in a resistor over a complete cycle is R1/2 I_(m)^(2)R (b) A light bulb is rated 100 W for a 220 V a.c. supply. Calculate the resistance of the bulb. |
|
Answer» Solution :(a) When an alternating current flows in the circuit is given by `I = I_(m) sin omega t` `therefore`AVERAGE power dissipated per COMPLETE cycle of a.c. is given by `P_(av) = 1/T int_(0)^(T) V I dt = 1/T int_(0)^(T) V_(m) sin omegat. I_(m) sin omega t dt = (V_(m) I_(m))/T int_(0)^(T) sin^(2) omega Tdt` `(V_(m)I_(m))/(2) = 1/2 I_(m)^(2)R [ therefore V_(m) = RI_(m))`] (b) Here P = 100 W, `V_(RMS) = 220V`. As `P= I_(rms)^(2) R = (V_(rms)^(2))/R` `rArr` Reactance `R = (V_(rms)^(2))/P = (220)^(2)/100 = 484 Omega` |
|