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?

Explanation

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`