1.

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»


Solution :
`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`


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