Question 717: Current Electricity

Question

In the circuit shown below, the key is pressed at time t = 0. Which of the following statement(s) is (are) true?
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l0pxasaa/e68f47e1-58e0-4a56-a1a4-2325a87bee78/6f656210-a327-11ec-8934-db0715b59417/file-1l0pxasab.png?format=png)

Accepted Answer

When the key is just pressed, there is no charge on capacitors
$	herefore$ $5 = 25000{I_1} \Rightarrow {I_1} = {5 \over {25000}} = 0.2\,mA$
and $5 = 50000{I_2} \Rightarrow {I_2} = {5 \over {50000}} = 0.1\,mA$
Reading of voltmeter $ = {V_Q} - {V_P} =  - 25000{I_1}$
$ =  - 25000 	imes 0.2 	imes {10^{ - 3}}V =  - 5V$
After a very long time, steady state is reached, i.e. ${I_1} - {I_2} = 0$
$	herefore$ $5 = {{{q_1}} \over {40 	imes {{10}^{ - 6}}}}$ or ${q_1} = 200\mu C$
and $	herefore$ $5 = {{{q_2}} \over {20 	imes {{10}^{ - 6}}}}$ or ${q_2} = 100\mu C$
Now, reading of voltmeter $ = {V_Q} - {V_P}$
$ = {{{q_2}} \over {20\mu F}} = {{100\mu C} \over {20\mu F}} = 5V$
For capacitor of 40$\mu$F,
${	au _1} = RC = 25 	imes {10^3} 	imes 40 	imes {10^{ - 6}} = 1\,s$
and for capacitor of 20$\mu$F,
${	au _2} = RC = 50 	imes {10^3} 	imes 20 	imes {10^{ - 6}} = 1\,s$
$	herefore$ At any instant t,
${q_1} = 200[1 - {e^{ - t}}]\mu C$, ${q_2} = 100[1 - {e^{ - t}}]\mu C$
${I_1} = 0.2\,{e^{ - t}}$ mA, ${I_2} = 0.1\,{e^{ - t}}$ mA
${V_Q} - {V_P} = {{100(1 - {e^{ - t}})\mu C} \over {20\mu F}} - 25k\Omega  	imes 0.2\,{e^{ - t}}$ mA
$ = 5(1 - {e^{ - t}}) - 5{e^{ - t}} = 5 - 10{e^{ - t}}$
At t = ln 2 s; ${V_Q} - {V_P} = 5 - 10{e^{ - \ln 2}} = 5 - {{10} \over 2} = 0$ V
Initially, I
0
= 0.1 mA + 0.2 mA = 0.3 mA
At t = 1 s, I = I
1
+ I
2
$ = (0.2{e^{ - 1}} + 0.1{e^{ - 1}}) = {{0.3} \over e}\,mA = {{{I_0}} \over e}$
After a long time, i.e. t $	o$ $\infty$
$I = {I_1} + {I_2} = 0.2{e^{ - \infty }} + 0.1{e^{ - \infty }} = 0$
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l0rat07x/d41cc533-a4c0-4ba5-8a42-0d062fde20a9/096b30d0-a3e9-11ec-9889-8ff85cf85963/file-1l0rat07y.png?format=png)

Answer

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

Answer

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

Answer

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

Answer

The current in the ammeter becomes zero after a long time

Current Electricity

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

Choose the correct answer: