InterviewSolution
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InterviewSolution
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(a) For a given a.c., `i = i_(m)` sin wt, show that the average power dissipated in a resistor R over a complete cycle is `(1)/(2) i._(m)^(2) R`. (b) A light bulb is rated at 100 W for a 220 V a.c. supply. Calculate the resistance of the bulb. |
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Answer» (a) The average power dissipated `bar(P)=(i^(2).R)=(i_(m)^(2)R sin^(2) omega t)=i_(m)^(2)R(sin^(2)omega t)` `therefore" "sin^(2)omega t=(1)/(2)(1-cos 2 omega t)" "=(sin^(2)omega t)=(1)/(2)(1-(cos 2 omega t))=(1)/(2)" "therefore" "(cos 2 omega t = 0)` `therefore" "bar(P)=(1)/(2)i_(m)^(2)R`. (b) Powet of the bulb, P = 100 W and voltage, V = 220 V. The resistance of the bulb is given as `R=(V^(2))/(P)=(220 xx 220)/(100)= 484 Omega`. |
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