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Electrostatic Potential And Capacitance

Question
CBSEENPH12038976
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(a) For a given a.c.,i space equals space i subscript m sin space omega t comma show that the average power dissipated in a resistor R over a complete cycle is 1 half i squared subscript m R

(b) A light bulb is rated at 100 W for a 220 V a.c. supply. Calculate the resistance of the bulb. 

Solution

a) Average power consumed in resistor R is given by, 
straight P subscript av space equals space fraction numerator 1 over denominator integral subscript 0 superscript straight T dt end fraction space integral subscript 0 superscript straight T straight i squared space straight R space dt space space space space space equals space fraction numerator straight i squared subscript straight m space straight R over denominator straight T end fraction space integral subscript 0 superscript straight T sin squared space ωt space dt space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis space space space space space space equals fraction numerator straight i squared subscript straight m space straight R over denominator 2 straight T end fraction integral subscript 0 superscript straight T left parenthesis 1 space minus space cos space 2 ωt right parenthesis space dt space space space space space space space space space space space space space space space space space space space space equals fraction numerator straight i squared subscript straight m space straight R over denominator 2 straight T end fraction space open square brackets integral subscript 0 superscript straight T dt space minus space integral subscript 0 superscript straight T cos space 2 ωt space dt close square brackets space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis thin space space space space space space space space equals fraction numerator straight i squared subscript straight m space straight R over denominator 2 straight T end fraction space left square bracket space straight T space minus space 0 right square bracket space space space space space space space space equals space fraction numerator straight i squared subscript straight m space straight R over denominator 2 straight T end fraction 
b) In case of ac, we have
straight P subscript av space equals space straight v squared subscript rms over straight R space equals space straight V squared subscript eff over straight R rightwards double arrow space straight R space equals space straight V squared subscript rms over straight P subscript av space space space space space space space space space equals space fraction numerator 220 space straight x space 220 over denominator 100 end fraction space space space space space space space space space equals space 484 space straight capital omega

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