In the circuit shown below, the key is pressed at time t = 0 . Which of the following statements (s) is (are) true?

Options

A. The voltmeter desplays `-5V` as soon as the key is pressed, and displays `+5V` after long time.
B. The voltmeter will display `0V` at time `t=ln2` seconds
C. The current in the ammeter becomes `1//e` of the initial value after 1 second
D. The current in the ammeter becomes zero after a long time.

Answer

A, B, C, D

Explanation

at `t=0` voltage across each capacitor is zero, so reading of voltmeter is `-5` volt. At `t=0`, capacitors are fully charged. So for ideal voltmeter, reading is 5 volt. At transistance state,
`I_(1)=(5)/(50)e^((l)/(tau))mA,I_(1)=(5)/(25)e^((l)/(tau))` and `I=I_(1)+I_(2)`
Where `tau=1` sec
So I becomes `1//e` times of the initial current after 1 sec.
The reading of voltmeter at any instant
`=DeltaV_(40muF)-DeltaV_(50kOmega)=5(1-e^(-(l)/(tau)))-5e^((l)/(tau))`
So at `t=ln2sec` reading of voltmeter is zero.