In the circuit shown in figure `R_(1) = R_(1) = 6R_(3) = 300M Omega,C = 0.01muF` and `E - 10V`. The switch is closed at `t =0`, find
(a) Charge on capacitor as a function of time
(b) enegry of the capacitor at `t = 20s`

Answer

(a) `q 0.05 (1-e^(-t//2))muC`; (b) `0.125 muJ`

Explanation

`R_(1) = R_(2) = 300M Omega, R_(3) = 50M Omega`
for loop `1`
`10 - 300 (I_(1) +I_(2)) - 300I_(1) = 0`
`I - 60I_(1) = 30I_(2) …(1)`
for loop `2`
`300I_(1) - 50I_(2) -(q)/(0.01) =0`
`6I_(1) - I_(2)- 2q = 0 ...(2)`
from `(1)` and `(2)`
`I -40I_(2) = 20q ...(3)`
Put `I_(2) = (dq)/(dt)` and integrating `q = (1)/(20)(1-e^(-t//2))`
time constant `tau = ((R_(1)R_(2))/(R_(1)+R_(2))+R_(3)) C = 2sec`
`q_(0) = (CER_(2))/(R_(1)+R_(2)) = 0.05muC`
`q = 0.05 (1-e^(-t//c)) rArr q=0.05 (1-e^(-t//2))`
(b) at `t = 20s = 10tau`, circuit behave like steady state condition
`V_(C ) = ((R_(2))/(R_(1)+R_(2))) xxR_(2) = 5V`
`U = (1)/(2)CV_(C)^(2) = 0.125muJ`