The plates of a capacitor of capacitance `10(mu)F`,charged to `60(mu)C`,are joined together by a wire of resistance `10(Omega)`at t=0.Find the charge on the capacitor in the circuit at `(a)t=0,(b)t=30(mu)s,(c )t=120(mu)s and (d)t=1.0ms`.

Explanation

`C = 10 mu f , Q = 60 mu C , R = 10 Omega `
` (a) at t = 0 , q = 60 mu C `
` (b) at t = 30 mu s , e^-0.3 = 44 mu c `
` q = 60 mu c `
` = 60 xx (10^-6) xx e^-0.3 = 44 mu c `
`(c) at t = 120 mu s `,
` q = 60 xx (10^-6) xx e^-1.2`
` = 18 mu c `
` (d) at t = 1 ms `
` q = 60 xx (10^-6) xx e^10 `
` = 0.00272 = 0.003 mu C ` .