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Show that the current in a pure resistor is in phase with the ac voltage across it and hence S.T. average power dissipation in a resistor is i^(2)R where I is r.m.s value of ac. |
Answer» SOLUTION : Let I be the current in a RESISTOR voltage drop a cross the resistor will be iR, supplying KVL to the closed loop. We write `v_(m)sinomegat=iR`. i.e., `i=((v_(m))/(R))sinomegat` i.e., `i=i_(m)sinomegat` where `i_(m)=(v_(m))/(R)` HENCE current is in phase with the APPLIED voltage. By definition of power `p=i^(2)R` i.e., `p=i_(m)^(2)Rsinomegat` Average power, `barp=(:i^(2)R:)=(:i_(m)^(2)Rsin^(2)omegat:)` Since, `(:sin^(2)omegat:)=(1)/(2)` `barp=i_(m)^(2)R(:sin^(2)omegat:)` i.e. `barp=(1)/(2)i_(m)^(2)R` 
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