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A capacitor charged to 10 V is being discharged through a resistance R. At the end of 1 s, the voltage across the capacitor is `5 V`. What will be the voltage after 2 s? |
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Answer» The equation of discharging of capacitor is `q = q_(0) e^(-t//RC) , (q_(0))/(q) = e^(t//RC)` or `t = RC log_(e) ((q_(0))/(q)) = RC log_(e) ((V_(0))/(V))` At `t = 1 s, V = 5` volt from `V_(0) = 10 V` `:. 1 = RC log_(e) ((10)/(5)) = RC log_(e) 2` At `t = 2 s, V = ? V_(0) = 10 V` `:. 2 = RC log_(e) ((10)/(V))` Dividing (i) by (i), we get `2 = (log_(e) 10 - log_(e) V)/(log_(e) 2)` `2 log_(e) 2 = log_(e) 10 - log_(e) V` `log_(e) V = log_(e) 10 - log_(e) 2^(2) = log_(e) ((10)/(4))` `V = (10)/(4) = 2.5 V` |
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