Only switch `S_1` closed in fig.
a. what is the steady-state reading of the voltmeter?
b. What is the charge on the `3muF` capacitor?
c. How much power does the 12V battery supply in the steady state after all the switches are closed?
d. What is the charge on `2muF` capacitor?

Explanation

a. As explained earliar that a capacitor is an open cirucit to DC
in steady state. The effective circuit through which current
will pass initially after closing switch `S_1` is shown is figure.
When steady state is reached, the capacitor will not allow
current. Thus, there is no current from battery. Reading of
voltmeter is zero.
b. Without current the resistor
act as conducting wires.
Hence, potentia difference
across capacitor is 12V. Charge on capacitor, `3muF,` is
`Q = (3xx2)muC = 36muC`

c. When all the switches are
closed, the branch contatining
capacitor will not allow
current after steady state is reached. The equivalent circuit
from point of view of current is redrawn as in figure.

Let the current in the single loop circuit be I. we begin at
left lower corner and traverse the circuit anticlockwise while
applying KVL. We get `-20 I +20-10 I - 12 = 0`
`I = (4)/(15)A`
Power received by 12 V battery is `P = VI = 12xx(4)/(15) = (16)/(5)W`
d. `V_c - 10 I - 12 = V_d`
or `V_c - V_d = (44)/(3)V or Q =CV = (88)/(3)muC`