In the circuit shown below,the key is pressed | vedclass

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(B) At $t=0$,capacitors are uncharged,so they act as short circuits. The voltmeter is connected across the junction of the two branches. Initially,the

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$A, B, C, D$

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(B) At $t=0$,capacitors are uncharged,so they act as short circuits. The voltmeter is connected across the junction of the two branches. Initially,the potential at the top node is $0 \ V$ and at the bottom node is $5 \ V$ (relative to the negative terminal),so the voltmeter reads $-5 \ V$. After a long time,capacitors are fully charged and act as open circuits. The circuit behaves as a voltage divider. The potential at the top node becomes $5 \ V$ and at the bottom node $0 \ V$,so the voltmeter reads $+5 \ V$. Thus,$(A)$ is true.The potential difference across the voltmeter $V_v = V_{top} - V_{bottom}$. Using the time-dependent charging equations for the $RC$ circuits,we find $V_v(t) = 5(2e^{-t} - 1)$. Setting $V_v = 0$ gives $2e^{-t} = 1$,or $t = \ln 2 \ s$. Thus,$(B)$ is true.The total current $I(t) = I_1(t) + I_2(t) = I_0 e^{-t/	au}$. The time constant for both branches is $	au = RC = (50 	imes 10^3 \Omega)(20 	imes 10^{-6} F) = 1 \ s$. Thus,$I(t) = I_0 e^{-t}$. At $t = 1 \ s$,$I = I_0/e$. Thus,$(C)$ is true.After a long time,capacitors are fully charged,acting as open circuits,so the current in the ammeter becomes zero. Thus,$(D)$ is true.Therefore,all statements $(A, B, C, D)$ are true.

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