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A capacitor, charged to 10 V , is being discharged through a resistance R. At the end of 1s the voltage across the capacitor is 5V. What will be the voltage after 2s ? |
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Answer» Solution :The equation for discharging of a capacitor through a resistance R is given by `q= q_0 E^(-t//CR)` or, `(q_0)/(q) = e^(t//CR)` or, `log_e (q_0)/(q) = t/(CR)` or , `t= CR log_e ((q_0)/(q))` As charge is proportional to voltage in case of a capacitor `t= CR log_e ((V_0)/(V))` Given that `V_0 = 10 ` volt At t = 1 sec, `V= 5 ` volt Hence `1= CR log_e ((10)/(5)) = CR log_e (2) ""....(1)` At t= 2 sec, the voltage will be given by : 2 = CR `log_e ((10)/(V)) "".....(2)` Dividing eq.(2) by eq. (1) , we get : `2= (log_e (10//V))/(log(2)) = (log_e 10- log_e V)/(log_e 2)` or , `log_e V= log_e 10-2 log_e 2` `=log_e 10 - log_e 4` or, `((10)/(4)) = (2.5) ` VOLTS |
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