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 current in the ammeter becomes zero after a long time.

At time $t=0$ or as soon as the key is pressed, voltmeter displays $-5V$ since the voltage across each capacitor is zero and after long time, capacitors will be fully charged and hence the voltmeter will display $5V$
Time constant for the circuit, $\tau =RC=40\times {10}^{-6}\times 25\times {10}^3=1sec$
The reading of voltmeter at any instant = $\Delta V_{40\mu F}-\Delta V_{50k\Omega }$
Voltage variation across capacitor, $\Delta V_{40\mu F}=V_{supply}(1-e^{\frac{-t}{\tau }})=5(1-e^{-t})$
Voltage variation across resistor, $\Delta V_{50k\Omega }=V_{supply}\left(e^{\frac{-t}{\tau }}\right)=5e^{-t}$
Reading of ammeter = $5(1-e^{-t})-5e^{-t}=5-10e^{-t}$
at $t=ln2$
Reading = $5-10e^{-ln2}=5-5=0$
Hence option 2 is correct.
Initially the capacitor gets short circuited and let the current to pass through it. Thus the equivalent circuit consists of both the resistor connected in parallel. Hence the equivalent resistance will be
$R_{eq}=\frac{50\times 25}{50+25}=\frac{50}{3}k\Omega$
Initial current across the the ammeter, $I_i=V/R=\frac{5}{50/3}mA$
Voltage variation with time across the resistors $25k\Omega$ and $50k\Omega$ can be given by $V=V_s{e}^{\frac{-t}{\tau }}=5e^{-1}V$
Current across $25k\Omega$, $I_1=V/R=5e^{-1}/25mA$
Current across $50k\Omega$, $I_2=V/R=5e^{-1}/50mA$
Total current across ammeter, $I=I_1+I_2=\frac{5}{50}\times 3e^{-1}=\frac{I_i}{e}$
Hence the current in the ammeter becomes $1/e$ of the initial value after $1$ second
Hence option 3 is correct.
Since the current in the ammeter at any instant is $I=\frac{5}{50}\times 3e^{-t}$
After long time $t=\infty$
Thus $I=\frac{5}{50}\times e^{-\infty }=0$
Hence the current in the ammeter becomes zero after long time.
Hence option 4 is also correct.