A capacitor of capacitance `100(mu)F` is charged by connecting it to a battery of emf 12 V and internal resistance `2(Omega)`. (a) Find the time constant of the circuit. (b) Find the time taken before 99% of maximum charge is stored on the capacitor.

Explanation

The time constant is ltbr.`(tau)=CR=(100(tau)F)(2(Omega)=200(tau)s.`
the charge at time t is
`q=(epsilon)C(1-e^((-t)/(CR))).`
Putting `q=0.99(epsilon)C,`
`0.99=1-e^((-t)/((200(mu)s))
,`-(t)/(200(mu))=1n(0.01)`
`or, t=920(mu)s=0.92ms.`