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

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The voltmeter will display 0 V at time $t = \ln 2$ seconds

The current in the ammeter becomes 1/e of the initial value after 1 second

The current in the ammeter becomes zero after a long time

The voltmeter displays -5V as soon as the key is pressed, and displays +5V after a long time

answer is A, B, C, D.

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Detailed Solution

$L e t V_{B} = 0 \Rightarrow V_{A} = 5 V$

Voltmeter is ideal ,thus no current will pass through it.
A) $A t t = 0$ q=0 in both the capacitor $V_{C} = V_{B} = 0$ and $V_{A} = V_{D} = 5 V$
$∴ V_{C} - V_{D} = - 5 V$
At $t = \infty$ : $I = 0 \Rightarrow V = 0$ across both resistors $V_{A} = V_{C} = 5 V$ $V_{D} = V_{B} = 0 V \Rightarrow V_{C} - V_{D} = 5 V$
This can be shown by voltmeter

B) $V_{A} - V_{D} = \frac{q_{1}}{C_{1}} = 5 [1 - e^{- t / R_{1} C_{1}}] = 5 [1 - e^{- \frac{t}{2}}]$
At $t = l n 2,$ $V_{A} - V_{D} = 5 [1 - \frac{1}{2}] = \frac{5}{2} V$
$V_{A} - V_{C} = E e^{\frac{- t}{R_{2} C_{2}}} = 5 e^{- \frac{t}{2}} = \frac{5}{2} V$
$(1) - (2) \Rightarrow V_{C} - V_{D} = 0$ reading of voltmeter