In a circuit shown in the figure, the capacitor 𝐶 is initially uncharged and the key 𝐾 is open. In this condition, a current of 1 A flows through the 1 Ω resistor. The key is closed at time 𝑡 = 𝑡0. Which of the following statement(s) is(are) correct? [Given: 𝑒 −1 = 0.36] (A) The value of the resistance 𝑅 is 3 Ω. (B) For 𝑡 < 𝑡0, the value of current 𝐼1 is 2 A. (C) At 𝑡 = 𝑡0 + 7.2 𝜇s, the current in the capacitor is 0.6 A. (D) For 𝑡 → ∞, the charge on the capacitor is 12 𝜇C.

 

Given:

1. The capacitor $C$ is initially uncharged.

2. The key $K$ is open at $t<t_0$, and a current of $1\text{A}$ flows through the $1\mathrm{\Omega }$ resistor.

3. The key $K$ is closed at $t=t_0$.

4. $e^{-1}=0.36$.

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Step 1: Analyze the circuit for $t<t_0$ (Key $K$ is open)

When the key $K$ is open, the capacitor is uncharged and acts as an open circuit. The circuit consists of the $1\mathrm{\Omega }$ resistor and the resistor $R$ in series with the voltage source $V$.

The current through the $1\mathrm{\Omega }$ resistor is given as $1\text{A}$. Using Ohm's Law:

$$V=I\cdot R$$

The voltage across the $1\mathrm{\Omega }$ resistor is:

$$V_{1\mathrm{\Omega }}=1\text{A}\times 1\mathrm{\Omega }=1\text{V}$$

The voltage across $R$ is:

$$V_R=V-V_{1\mathrm{\Omega }}=V-1\text{V}$$

The current through $R$ is the same as through the $1\mathrm{\Omega }$ resistor, which is $1\text{A}$. Therefore:

$$R=\frac{V_R}{I}=\frac{V-1}{1}=V-1\mathrm{\Omega }$$

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Step 2: Analyze the circuit for $t\ge t_0$ (Key $K$ is closed)

When the key $K$ is closed, the capacitor starts charging through the resistors. The time constant $\tau$ of the RC circuit is:

$$\tau =R_{\text{eq}}\cdot C$$

where $R_{\text{eq}}$ is the equivalent resistance seen by the capacitor. Assuming $R_{\text{eq}}=R+1\mathrm{\Omega }$, the time constant is:

$$\tau =(R+1)\cdot C$$

The current in the capacitor as a function of time is:

$$I_C(t)=I_0\cdot e^{-t\mathrm{/}\tau }$$

where $I_0$ is the initial current through the capacitor at $t=t_0$.

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Step 3: Evaluate the statements

(A) The value of the resistance $R$ is $3\mathrm{\Omega }$.

From the analysis for $t<t_0$, we have:

$$R=V-1\mathrm{\Omega }$$

If $R=3\mathrm{\Omega }$, then:

$$V=R+1=3+1=4\text{V}$$

This is consistent with the given information. Therefore, statement (A) is correct.

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(B) For $t<t_0$, the value of current $I_1$ is $2\text{A}$.

The current through the $1\mathrm{\Omega }$ resistor is given as $1\text{A}$, not $2\text{A}$. Therefore, statement (B) is incorrect.

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(C) At $t=t_0+7.2\mu \text{s}$, the current in the capacitor is $0.6\text{A}$.

The time constant $\tau$ is:

$$\tau =(R+1)\cdot C$$

If $R=3\mathrm{\Omega }$ and $C$ is such that $\tau =7.2\mu \text{s}$, then:

$$\tau =(3+1)\cdot C=4C=7.2\mu \text{s}$$

Thus:

$$C=\frac{7.2\mu \text{s}}{4}=1.8\mu \text{F}$$

The current in the capacitor at $t=t_0+7.2\mu \text{s}$ is:

$$I_C=I_0\cdot e^{-t\mathrm{/}\tau }=I_0\cdot e^{-1}=I_0\cdot 0.36$$

If $I_0=1\text{A}$, then:

$$I_C=1\cdot 0.36=0.36\text{A}$$

This is not equal to $0.6\text{A}$. Therefore, statement (C) is incorrect.

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(D) For $t\to \mathrm{\infty }$, the charge on the capacitor is $12\mu \text{C}$.

As $t\to \mathrm{\infty }$, the capacitor is fully charged, and the voltage across it is equal to the total voltage $V$. The charge on the capacitor is:

$$Q=C\cdot V$$

If $V=4\text{V}$ and $C=1.8\mu \text{F}$, then:

$$Q=1.8\mu \text{F}\cdot 4\text{V}=7.2\mu \text{C}$$

This is not equal to $12\mu \text{C}$. Therefore, statement (D) is incorrect.

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Final Answer:

Only statement (A) is correct. The correct option is:

$$\boxed{{\displaystyle A}}$$