InterviewSolution
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InterviewSolution
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At `t=0` switch `S` is closed . (a) After how much time , charge on capacitor is 50% of maximum charge stored on capacitor. (b) After how much time, energy stored in capacitor is halt of maximum energy stored in capacitor. |
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Answer» (a) `q = CE (1-e^(-t//RC))` `q_(max) = CE, q = (q_(max))/(2) = 1/2 CE` `1/1 CE = CE(1-e^(-t//RC))` `e^(-t//RC)=2` `t/(RC)`log_(e) e = log_(e) 2 = 0.693RC` (b) `U =(q^2)/(2C) = 1/2 CE^(2)(1-e^(-t//RC))^(2)` `U_(max) = 1/2 CE^2, U = (U_(max))/(2) = (CE^2)/(4)` `(CE^2)/(4) = (CE^2)/() (1-e^(-t//RC))^(2)` `(1-e^(-t//RC))^(2) = 1/2 implies (1-e^(-t//RC))=1/(sqrt(2))` `e^(-t//RC)=1-1/(sqrt(2)) = (sqrt(2)-1)/(sqrt(2))` `e^(-t//RC) = (sqrt(2))/(sqrt(2)-1) = (sqrt(2))/(sqrt(2)-1)* ((sqrt(2)-1))/((sqrt(2)+1)) = (2+sqrt(2))` `t/(RC ) log_(e) e = log_(e) (2+sqrt(2))` ltbr .`t = RC log_(e) (2+sqrt(2))`. |
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