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]

Correct Answer: All four statements are correct:

• The value of the resistance R is 3 Ω.
• For t < t₀, the value of current I₁ is 2 A.
• At t = t₀ + 7.2 μs, the current in the capacitor is 0.6 A.
• For t → ∞, the charge on the capacitor is 12 μC.

Step 1: Circuit analysis for t < t₀ (Key K is OPEN)

When key K is open, no current flows through the bottom-most branch containing capacitor C = 2 μF and the 3 Ω resistor. Therefore, the circuit effectively consists of three parallel branches:

1. Top branch: Battery of 15 V with resistance R.
2. Middle-1 branch: Battery of 5 V with resistor 1 Ω.
3. Middle-2 branch: Resistor 3 Ω carrying current I₁.

Let the potential of the common right node be 0 V, and let the potential of the common left node (before key K) be V_A.

Using Nodal Analysis at node A:

$\frac{15-V_A}{R}=\frac{V_A-5}{1}+\frac{V_A}{3}$

We are given that the current through the 1 Ω resistor is 1 A. Current flows from the 5 V source towards node A through the 1 Ω resistor, so:

$\frac{5-V_A}{1}=1\text{A}\Rightarrow V_A=6\text{V}$

Now, substituting V_A = 6 V into our nodal equation:

$\frac{15-6}{R}=\frac{6-5}{1}+\frac{6}{3}$

$\frac{9}{R}=1+2=3\Rightarrow R=3\text{Ω}$

The current I₁ flowing through the 3 Ω resistor is:

$I_1=\frac{V_A}{3}=\frac{6}{3}=2\text{A}$

Thus, statements 1 and 2 are correct.

Step 2: Circuit analysis after closing Key K at t = t₀

To analyze the charging of the capacitor, we convert the network connected across the capacitor branch (between node A and the right node) into its Thévenin Equivalent Circuit.

1. Thévenin Equivalent Voltage (V_th):
The open-circuit potential difference across node A and the right node when no current flows into the capacitor branch is:

$V_{\text{th}}=V_A=6\text{V}$

2. Thévenin Equivalent Resistance (R_th):
Deactivating independent voltage sources gives three parallel resistors: R = 3 Ω, 1 Ω, and 3 Ω.

$\frac{1}{R_{\text{th}}}=\frac{1}{3}+\frac{1}{1}+\frac{1}{3}=\frac{5}{3}\Rightarrow R_{\text{th}}=0.6\text{Ω}$

The total resistance connected in series with the capacitor C = 2 μF is:

$R_{\text{total}}=R_{\text{th}}+3\text{Ω}=0.6+3=3.6\text{Ω}$

Step 3: Calculating Time Constant (τ)

$\tau =R_{\text{total}}\times C=3.6\text{Ω}\times 2\mu \text{F}=7.2\mu \text{s}$

Step 4: Current in the capacitor at t = t₀ + 7.2 μs

The initial current in the capacitor at t = t₀⁺ (since initially uncharged, V_C = 0):

$I_0=\frac{V_{\text{th}}}{R_{\text{total}}}=\frac{6}{3.6}=\frac{5}{3}\text{A}$

The charging current decay equation is:

$I\left(t\right)=I_0{e}^{-(t-t_0)/\tau }$

At t = t₀ + 7.2 μs (which corresponds to t - t₀ = τ):

$I=\frac{5}{3}\times e^{-1}=\frac{5}{3}\times 0.36=0.6\text{A}$

Thus, statement 3 is correct.

Step 5: Charge on the capacitor for t → ∞

In steady-state (t → ∞), the capacitor is fully charged and acts as an open circuit. The voltage across the capacitor equals the Thévenin voltage V_th = 6 V.

$Q=C\times V_{\text{th}}=2\mu \text{F}\times 6\text{V}=12\mu \text{C}$

Thus, statement 4 is also correct.