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 displays `(-5V) as soon as the key is pressed, and displays `(+5 V)` after a long time
B. The volmeter will display 0V at time t = In 2 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

(a,b,c,d)
At t = 0 , Capacitors act as open circuit and voltmeter display
`(-5v)`
At `t= prop`, Capacitor acts as open circuit and no current flows through voltmeter(:. Very high resistance of
voltmeter)
So it display +5V. (a) is the correct option.
`q_1 = 2CV (1-e^(-t//2CR)), i_1 = V/R e^(-t/(2CR))`
`q_2 = CV(1-e^(-t/(2CR))), i_2 = V/(2R) e^(-t/(2CR))`

`:. Delta V = -i_2 xx 2R + (CV_1)/(2C) = V[1-2e^(-t//(2CR))] = 0`
(b) is the correct options
`At tau = 1 sec, i = i_(0)/e` [:. i = i_0 e^(-t/(tau))]`
`:. (c) is the correct option
After a long time no current flows since both capacitor
and voltmeter do not allow current to flow.
`:. (d) is the correct option.