At `t=0` switch `S` is closed . Find
`(a) charge on capacitor after one time constant
(b) the time when p.d. across resistance equals p.d. across capacitance
(c ) maximum energy stored in capacitor.

Explanation

(a) We know, charge on capacitor at any time `t`
`q = CE (1-e^(-t//RC))`
`t = tau = RC`
`q = CE(1-e^(-RC//RC))=CE(1-e^(-1))=CE(1-1//e)`
`e = 2.7`
`q = CE (1-1/2.7) = 0.63CE`
(b) p.d. across capacitor
`V_(C) = q/C = E(1-e^(-t//RC))` ..(i)
p.d. across resistor
`V_R= E e^(-t//RC)` ..(ii)
`{V_R+V_C=E}`
`E e^(-t//RC) = 1 implies e^(-t//RC) = 1/2 implies e^(-t//RC) = 2`
Taking log on both sides on the base `e`
`t/(RC) log_(e)e = log_(e)2`
`t/(RC) log_(e) e = log_(e)2`
`t/(RC) = log_(e)2 implies t = RC Log_(e) 2 = 0.693 RC`
(c) energy stored in capacitor
`U = (q^2)/(2C) = 1/2 CE^(2) (1-e^(-t//RC^2))`
`U ` is maximum when `t = oo`
`U_(max) = 1/2 CE^(2)`.