1)

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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A) The voltmeter display -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:

Option A,B,C,D

Explanation:

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Just after pressing key,

  5-25000i1 =0

5-50000 i2 =0

 ( As charge in both capacitors =0)

i1=0.2mAi2=0.1mA

And  VB+25000i1=VA

VBVA=5V

After a long time , i1  and i2 =0 ( steady state)

5q140=0q1=200μC

and  5q220=0q2=100μC

VBq220=VAVBVA=+5V

(a) is correct.

For capacitor 1,

q1=200[1etl]μCi1=15etlmA

For capacitor 2,

q2=100[1etl]μCi2=110etlmA

VBq220+i1×25=VA

VBVA=5[1et]5et

     =5[12et]

 At t=ln 2,

  VBV=5[11]=0

(b) is correct.

At t=1,

 i=i1+i2=15e1+110e1=310.1e

At t=0

i=i1+i2=15+110=310

(c) is correct

 After a long time , i1=i2=0

(d) is correct.