Question

In the circuit shown below, the key is pressed at time t = 0. Which of the following statement(s) is (are)
true?
(A) The voltmeter displays – 5 V as soon as the key is pressed, and displays +5 V 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

Given that,

The key is pressed at time t = 0.

According to figure,

The instantaneous equation, with q₁ being charge on upper and q₂ being that on lower capacitor

The equation of charge for upper capacitor,

$q'=q(1-e^{-t})$

Put the value of q

$q_{1}=40\times10^{-6}\times5(1-e^{-t})$

$q_{1}=200\times10^{-6}(1-e^{-t})$

The equation of charge for lower capacitor,

$q_{2}=20\times10^{-6}\times5(1-e^{-t})$

$q_{2}=100\times10^{-6}(1-e^{-t})$

We need to calculate the value of t

Using ohm's law

$V=IR$

$\dfrac{q_{1}}{C}=R\dfrac{dq_{2}}{dt}$

Put the value into the formula

$\dfrac{200\times10^{-6}(1-e^{-t})}{40\times10^{-6}}=50\times10^{3}\dfrac{d}{dt}(100\times10^{-6}(1-e^{-t}))$

$\dfrac{200\times10^{-6}(1-e^{-t})}{40\times10^{-6}}=50\times10^{3}\times100\times10^{-6}e^{-t}$

$10e^{-t}=5$

$e^{-t}=0.5$

$t=ln(2)$

We need to calculate the total current in circuit

Using formula of total current

$I=I_{1}+I_{2}$

$I=\dfrac{dq_{1}}{dt}+\dfrac{dq_{2}}{dt}$

Put the value into the formula

$I=\dfrac{d}{dt}(200\times10^{-6}(1-e^{-t}))+\dfrac{d}{dt}(100\times10^{-6}(1-e^{-t}))$

$I=200\times10^{-6}e^{-t}+100\times10^{-6}e^{-t}$

$I=100\times10^{-6}(2e^{-t}+e^{-t})$

$I=0.0003e^{-t}$

If t = ∞ then,

$I=0.0003e^{-\infty}$

$I=0$

According to solution,

All statements are true

Hence, This is required solution.

All option is correct.