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 $-5V$ as soon as the key is pressed, and displays $+5V$ after a long time
• B. The voltmeter will display $0V$ 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)

at instant $t$ - (here ammeter & voltmeter are ideal)

$C_1=40\mu F,R_1=25k\Omega$
$C_2=20\mu F,R_2=50k\Omega$
$i_1=\frac{E}{R_1}{e}^{-\frac{t}{R_1{C}_1}}$
i.e. $i_1=\left(0.2e^{-1}\right)mA$
$\&i_2=\left(0.1e^{-1}\right)mA$
$i=i_1+i_2=\left(0.3e^{-t}\right)mA$
$i=0.3e^{-t}mA$
at $t=\infty ,i=0$ i.e. option (D)
at $t=1,i=\frac{0.3}{e}=\frac{i_0}{e}$ i.e. option (C)
at $t=0$ (just after key is pressed)- Reading of voltmeter
$V=V_N-V_M=V_B-V_A=-5$ volt
at $t=\ell n2\sec$
$i_1=0.2e^{-\ln 2}$
$i_1=0.1mA$
and $i_2=0.05mA$
Apply KVL from $N\to M$ via cell
$V_N+50\times 0.05-5+25\times 0.1-V_M=0$
$V_N-V_M=0$
i.e. Reading of voltmeter is zero i.e. option (B) at $t=\infty ,i=0$
hence reading of voltmeter, $V=V_N-V_M$
$V=5V$