A capacitor charged to `100V` is discharged by connecting the two ptates at `t=0` . If the potential difference across the plates drops to `1.0V` at `10 ms`, what will be the potential difference at `t= 20 ms`.

A capacitor charged to `100V` is discharged by connecting the two ptat

1.	A capacitor charged to `100V` is discharged by connecting the two ptates at `t=0` . If the potential difference across the plates drops to `1.0V` at `10 ms`, what will be the potential difference at `t= 20 ms`.
Answer» In discharging `RC` circuit charge on capacitor `q = q_(0)r^(-1//RC) = CV_(0)e^(-t//RC), V_(0)=100V` p.d. across capacitor `V_(C) = q/C = V_(0)e^(-t//RC)` `1 = 100 e^(-10xx10^(-3))//RC) implies e^((-10^(-2))/(RC)) = 1/100` At `t = 20 ms` `V_(C) = V_(0)e^(-t//RC)= 100e^((-20xx10^(-3))/(RC) = 100e^((-2xx10^(-2))/(RC))` `=100(e^((-10^(-2))/(RC))^(2) = 100 xx 1/((100)^(2)) = 0.01 V`.

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A capacitor charged to `100V` is discharged by connecting the two ptates at `t=0` . If the potential difference across the plates drops to `1.0V` at `10 ms`, what will be the potential difference at `t= 20 ms`.

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