An uncharged capacitor of capacitance is connected in a circuit with two ideal batteries and two ideal ammeters G and `G_(1)` as shown. Both switches are initially open. Each option below details a possible pattern of closing the switches, and the consequent readings of the ammeters. Choose the correct option(s):

Options

A. `S_(1)` is closed. After a long time interval, if `S_(2)` is closed while keeping `S_(1)` closed, initially G reads 1 A, and `G_(1)` reads zero
B. `S_(1)` is closed. After a long time interval, `S_(2)` is closed while keeping `S_(1)` closed. After another long time interval, G reads zero, and `G_(1)` reads 1.5A
C. Both `S_(1)` and `S_(2)` are closed simultaneously. Initially G reads 3 A, and `G_(1)` reads 2 A
D. Both `S_(1)` and `S_(2)` are closed simultaneously. After a long time interval, G reads zero, and `G_(1)` reads 0.5 A

Answer

A, C, D

Explanation

Let the left and right junction be called P and Q respectively,
Case I: `S_(1)` is closed.
At steady state, Current through the capacitor has become zero and therefore, `V_(P)-V_(Q)=20V`. When `S_(2)` is closed, the potential difference across the capacitor cannot change immediately. So, initially `V_(P)-V_(Q)` remains 20V. Therefore using this we can deduce that initially, current through `S_(1)` is zero and current through `S_(2)` is 1A from right towards left.
Case II: `S_(1) and S_(2)` are closed simultaneously.
Initially, the potential difference across the capacitor must be zero since it has no charge on it. hence initially, `V_(P)-V_(Q)=0`. using the we can find that initally, current through `S_(1)` is 2A from left towards right and current through `S_(2)` is 1 A from left towards right. therefore, by using KCL, we can find that current through the branch containing the capacitor is 3A from right towards left.