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Current Electricity

Section: Physics

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Question : 10 of 29

Marks: +1, -0

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

[JEE Adv 2016 P2]

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

Solution:

(A) At t = 0,capacitor acts as short-circuit

At

t \to \infty

,capacitor acts as open circuit & no current flows through voltmeter.

(B)

q_x=2CV\left(1-e^{\frac{-t}{2CR}}\right)\quad x=\frac{V}{R}e^{\frac{-t}{2CR}}

q_y=CV\left(1-e^{\frac{-t}{2CR}}\right)\quad y=\frac{V}{2R}e^{\frac{-t}{2CR}}

\Delta V=-y2R+\frac{q_x}{2C}

=V\left[1-2e^{\frac{-t}{2CR}}\right]=0

(C) τ= 1 sec

So by

i=1_0e^{\frac{-t}{\tau}}

current at t = 1 sec is

= \frac{i_0}{e}

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