For the circuit arrangement shown in fig.

a. Find the potentail difference across each capacitor in the steady - state condition.
b. Find the current through the `60Omega` resistor just after the instant when the key K is opened.

Explanation

In the steady state, no current passes through upper three
resistors. So `I = (12)/(10+30) = 0.3A`

Potential difference between A and B is `V =30xx0.3 = 9V.`
Now in loop ACDBA:
`(q)/(C_(1)) +(q)/(C_(2)) = 9`
or `(q)/(20) +(q)/(10) = 9 or q = 60muC`
a. Potential difference across `C_(1) :`
`V_(1) = (q)/(C_(1)) =(60)/(20) = 3V`
Potential difference across `C_(2) :`
`V_(2) = (q)/(C_(2)) = (60)/(10) = 6V`
b. After K is opened, 12 V will be out of circuit , and capacitors
will act as batteries as shown in fig.

Now one can find `I_(1) = 0.0375A.`