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

Explanation

Charging current `i=(V_0)/(R ) e^((t)/(RC))`
`therefore` Potential difference across R is
`V_R = iR = V_0 e^(-(t)/(RC))`
`therefore` Potential difference across .C. is `V_C = V_0 - V_R`
`= V_0 - V_0e^((t)/(RC)) = V_0 (1-e^((t)/(RC)) )`
but given `V_C = 3V_R` , we get
`1-e^(-t//RC) = 3e^(-t//RC) or 1 = 4e^(-t//RC)`
`e^((t)/(RC)) = 4 rArr (t)/(RC) = ln 4 rArr 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