In the circuit shown below, the key is pressed at time $t = 0$ . Which of the following statement(s) is(are) true [2016]

Question

In the circuit shown below, the key is pressed at time $t = 0$ . Which of the following statement(s) is(are) true? [2016]

Smart Achievers · Accepted Answer

(1, 2, 3, 4)

At $t = 0$ , i.e., as soon as the key is pressed, capacitors act as short circuit and voltmeter displays reading $- 5 V$ .

At $t = \infty$ , i.e., after a long time of key press, capacitors act as open circuit and no current flows through voltmeter ( $∵$ very high resistance of voltmeter) so it displays $+ 5 V .$

$q_{1} = 2 C V (1 - e^{- t / 2 C R}),$ $i_{1} = \frac{V}{R} e^{- \frac{t}{2 C R}}$

$q_{2} = C V (1 - e^{- t / 2 C R}),$ $i_{2} = \frac{V}{2 R} e^{- \frac{t}{2 C R}}$

$∴$ $Δ V = - i_{2} (2 R) + \frac{C V_{1}}{2 C}$ $= V [1 - 2 e^{- t / 2 C R}] = 0$

i.e., at time $t = \ln 2$ s voltmeter will display reading 0V.

$At τ = 1 s, i = \frac{i_{0}}{e}$ $[∵ i = i_{0} e^{- t / τ}]$

i.e., after 1s current in the ammeter becomes $\frac{1}{e}$ of the initial value.

After a long time no current flows since both capacitor and voltmeter do not allow current to flow.

Solution

Q.
In the circuit shown below, the key is pressed at time $t = 0$ . Which of the following statement(s) is(are) true [2016]

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

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

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

4 The current in the ammeter becomes zero after a long time.

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