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

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)` or `1=4e^(-t//RC)` `e^((t)/(RC))=4 implies (t)/(RC)=ln4 implies t= 2RC ln2`
`t = 2.5 xx 10^(6)xx 4 xx 10^(-6)xx2 xx 0.693` or t = 13.86 sec