In the circuit shown, `E = 18 kV, C = 10 muF,R_1 = 4 MOmega, R_2 = 6 MOmega , R_3 = 3MOmega` with `C` completely uncharged, switch S is suddenly closed (at `t = 0`).

(a) Determine the current through each resistor for `t=0` and `t=oo`
(b) What are the values of `V_2` (potential difference across `R_2`) at `t = 0` and `t = oo`. ?
(c) Plot a graph of the potential difference` V_2` versus `t` and determine the instantaneous value of `V_2`.

Explanation

At `t=0`, when capacitor is uncharged, its equivalent resistance is zero.
`:. R_(n et)=4+(6xx3)/(6+3)=6MOmega`
or `i_1=(18xx10^3)/(6xx10^6)A=3mA`
This will distribute in inverse ratio of resistance,
`:. i_2=3/(6+3)i_1=1mA` and `i_3=2mA`
At `t=oo` when capcitor is completely charged, equivalent resistance of capacitgor is infinite
`i_3=0,i_1=i_2=(18xx10^3)/((4+6)xx10^6)=1.8mA`
b. At `t=0`
`V_2=i_2R_2=(1xx10^-3)(6xx10^6)V`
`=6kV`
At `t=oo`
`V_2=i_2R_2=(1.8xx106-3)(6xx10^6)V`
`=10.8kV`
c. to find time constant of the circuit we will have to short circuit the battery and find resistance across capacitor. In that case `R_1` and `R_2` are in parallel and they are in series with `R_3`.
`:. R_(n et)=3+(4xx6)/(4+6)=5.4MOmega`
`:. tau_C=CR_(net)=(10xx10^-6)(5.4xx10^6)`
`=54s`
`:. V_2=6+(10.8-6)(1-e^(-t/C))`
`=6+4.8(1-e^(t/54))`
Here, `V_2` is in `kV` and `t` is second.