InterviewSolution
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InterviewSolution
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A `4 muF` capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged `2 mu F` capacitor. How much electrostatic energy of the first capacitor is disspated in the form of heat and electromagnetic radiation ? |
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Answer» Here, `C_(1) = 4 muF = 4xx10^(-6) F, V_(1) = 200 vol t` Initial electrostatic energy stored in `C_(1) si U_(1) = (1)/(2) C_(1) V_(1)^(2) = (1)/(2) xx 4xx10^(-6)xx200xx200 = 8xx10^(-2) joul e` When `4 mu F` capacitor is connected to uncharged capacitor of `2 mu F`, charge flows and both acquire a common potential `V = ("total charge")/("total capacity") , V = (C_(1) V_(1))/(C_(1) + C_(2)) = (4xx10^(-6)xx200)/((4+2) 10^(-6)) = (800)/(6) vol t` `:.` Final electrostatic energy of both capacitors `U_(2) = (1)/(2) (C_(1) + C_(2)) V^(2) = (1)/(2) xx6xx10^(-6) xx (800)/(6) xx (800)/(6) = 5.33xx10^(-2) joul e` `:.` Energy dissipated in the form of heat and electromagnetic radiation `U_(1) - U_(2) = 8xx10^(-2) - 5.33xx10^(-2) = 2.67xx10^(-2) joul e` |
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