A capacitor `C_(1)` is charged to a potential V and connected as shown with capacitors `C_(2) and C_(3)` both of which are initially uncharged. Capacitance of all capacitors is C. The ammeter is ideal. At `t=0_(1)` switch S is closed. Pick the correct choice(s):

Options

A. Just after t = 0, the ammeter reads `(2V)/(3R)`
B. Just after t = 0, rate of total heat dissipation in all resistors is `(2V^(2))/(3R)`
C. After a long time, potential difference across `C_(1) is V/3`
D. If switch is kept closed for a long time, Total heat dissipation in all resistors is `(1)/(6)CV^(2)`

Answer

A, B, C

Explanation

Due to the symmetrical situation of `C_(2) and C_(3)`, current
into them is always equal.
At t=0, `C_(2) and C_(3)` are uncharged
So, applying KVL,
`V-iR-2iR=0 implies` At `t=0, i=(V)/(3R)`
`implies` Current out of `C_(1)=2i=(2V)/(3R),` Rate of heat dissipation `=i^(2)R+i^(2)R+(2i)^(2)R=6i^(2)R=(2V^(2))/(3R)`
At steady state, each capacitor curries charge `(CV)/(3)` with all left plates carrying positive charge.
`implies`Toal heat dissipation`=U_(i)-U_(f)=(1)/(2)CV^(2)-3((1)/(2)C((V)/(3))^(2))=(1)/(3)CV^(2)`.