Question

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

• A. The voltmeter displays $-5V$ as soon as the key is pressed, and displays $+5V$ after a long time
• B. The voltmeter will display 0 V at time $t=ln$ 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 1

Answer: (A), (B), (C), (D)

The correct options are
A The voltmeter displays $-5V$ as soon as the key is pressed, and displays $+5V$ after a long time
B The voltmeter will display 0 V at time $t=ln$ 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.

Writing instantaneous equation, with $q_1$ being charge on upper and $q_2$ being that on lower capacitor,

$q_1=(200\times {10}^{-3})(1-e^{-t})$

$q_2=(100\times {10}^{-3})(1-e^{-t})$

$\frac{q_1}{C}=(50\times {10}^3)\frac{d{q}_2}{dt}$

$\frac{200\times {10}^{-3}(1-e^{-t})}{40\times {10}^{-6}}=(50\times {10}^3)(100\times {10}^{-3})e^{-t}$

$e^{-t}=0.5$

$t=ln\left(2\right)$

$I=I_1+I_2=(200\times {10}^{-3})\left(e^{-t}\right)+(100\times {10}^{-3})e^{-t}$

$=100\times {10}^{-3}[2e^{-t}+e^{-t}]$

$=0.3e^{-t}$

At $t=\infty ,I=0$

Answer is A,B,C and D.