Question

In the circuit shown below, if at time $t=0$ the switch (key) is on, then which of the following statements is/are true?

• A. The voltmeter displays $-5\text{V}$ as soon as the key is pressed, and displays $+5\text{V}$ after a long time.
• B. The voltmeter will display $0\text{V}$ at time $t=\ln 2\text{s}$.
• C. The current in the ammeter becomes $\frac{1}{e}$ of the initial value after $1\text{s}$
• 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 $t=0$, capacitor acts as short circuit. So voltmeter displays $-5\text{V}$.

At time, $t=\infty$. Capacitor acts as open circuit, so voltmeter displays $+5\text{V}$

At $t=\infty$


Since, ideal voltmeter has infinite resistance current through it is zero. Current across ammeter is measured zero.

So, options (a) and (d) are correct.

Time constant of upper $\text{RC}$ circuit.

${\tau }_1=R_1{C}_1=25\times {10}^5\times 40\times {10}^{-6}=1\text{s}$

Time constant of lower, $\text{RC}$ circuit,

${\tau }_2=R_2{C}_2=50\times {10}^3\times 20\times {10}^{-6}=1\text{s}$

At time $t$:


$I_1=I_0{e}^{-\frac{t}{{\tau }_1}}$

$\Rightarrow I_1=\frac{5}{R_1}{e}^{-t}$

Similarly , $I_2=\frac{5}{R_2}{e}^{-t}$

Charge , $q_1=C_{1}V(1-e^{-\frac{t}{{\tau }_1}})$

$\Rightarrow q_1=5C_1(1-e^{-t})$

Similarly, $q_2=5C_2(1-e^{-t})$

Ammeter reading , $I=I_1+I_2$

$\Rightarrow I=5e^{-t}(\frac{1}{R_1}+\frac{1}{R_2})$

At $t=0\text{sec}$,

Initial current , $I_0=\frac{5}{R_1}+\frac{5}{R_2}$

At $t=1\text{sec}$ ,

Initial current , $I_0=\frac{1}{e}(\frac{5}{R_1}+\frac{5}{R_2})$

So, option (c) is the correct answer.

At $t=\ln 2$

Potential difference between $\left(3\right)$ and $\left(1\right)$,

$V_{31}=5(1-\frac{1}{2})=\frac{5}{2}\text{V}$

Potential difference between $\left(3\right)$ and $\left(2\right)$,

$V_{32}=I_2{R}_2=5\left(e^{-t}\right)$

At $t=\ln 2$

$V_{32}=\frac{5}{2}\text{V}$

So, voltmeter reading will be zero at $t=\ln 2\text{sec}$.

So, option (b) is correct

Hence, all the options are the correct answers.