Condsider
.
The charge on capacitor `C_(1)` is.

Options

A. If key `S_(1)` kept closed for long time such that capacitors are fully charged, the voltage difference between points and will be 10 V.
B. The key `S_(1)` is kept closed for long time such that capacitors are fully charged. Now key `S_(2)` is closed at this time, the instantaneous current across `30Omega` (between points P and Q) will be 0.2 A (round off to 1st decimal place).
C. At time `t=0`, the key `S_(1)` is closed, the instantaneous current in the closed circuit will be
D. If key `S_(1)` is kept closer for long time such that capacitors are fully charged, the voltage across the capacitor `C_(1)` will be 4V.

Answer

C, D

Explanation

just after `S_1` is closed,capacitor behave as zero resistance wires ( as they are initially uncharged) `thereforeI=(5)/(100+30+70)A=25mA therefore` ( C ) is correct
After `S_1` is closed for a long time current will be zero,Assuming total charge circulation to be q in anticlockwise direction,and applying KVL,
`-q/10-q/80+5-q/80=0 implies q=40 muC`
`therefore` Voltageacross `C_1=40/10=4` volts `therefore` (D) is correct.
Also `V_p-V_Q=V_(C_1)=4` Volts `therefore` (1) is wrong
NOw immediately after `S_2` is also closed,charge remin the same
`therefore V_(C_1)=40/10=4V, V_(C_2)=0 , V_(c_2)=40/80=0.5v, V_(C_V)=40/80=0`
Appllying KVL in two loops
`4+30xx-10+70(x+y)=0 implies 100x+70y=6`
`4+0.5-5+30y+0.5+100y+70(x+y)=0`
`200y +70x=0` .......(ii) Solving (i) and (ii) ,x=0.0769 A.