An uncharged capacitor is connected to a 15 V battery through a resistance of `10Omega.` It is found that in a time fo `2mus,` potential difference across the capacitor. Take in (1.5) = 0.4.

Explanation

We know that charge on the capacitor at any tieme is given
by ` q= Q(1 -e^(-t//pi))` where `Q = EC -15C`. Here cahrge q at any
time is given by `q =VC` where V is potential difference across
capacitor at that time. Here `V = -5 V`, so `q= 5C`, putting the
values, we get
`5C =15C(1-e^(-t//pi)) or e^(-t//pi) = 2//3`
or` (t)/(pi) = In((3)/(2)) or (t)/(Rc) =In ((3)/(2))`
or `C =(t)/(R In (3//2)) = 0.5 muF`