A `4 muF` capacitor, a resistance of `2.5 M Omega` is in series with `12V` battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [ Given In (2) = 0.693]

Explanation

`{:a)` Charging current `i = (V_(0))/(R )e^((t)/(RC))`
`:.` Potential difference across `R` is
`V_(R)-iR=V_(0) e^(-t/(RC))`
`:.` Potential difference across 'C' is `V_(C)=V_(0)-V_(R)`
`V_(0)-V_(0)e^(-(t)/(RC)) = V_(0)(1-e^(-(t)/(RC)))`
but given `V_( c) = 3V_(r )`, we get
`1-e^(-t//RC) = 3e^(-t//RC)` or `4e^(-t//RC)`
`e^((-t)/(RC)) = 4 rArr (t)/(RC) = ln 4 rArr t = t = 2RC ln 2`
`t = 2.5 xx 10^(6) xx 4 xx 10^(-6) xx 2 xx 0.693`
or `t = 13.86 sec`