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\; resistor. The key is closed at time t=t₀. Which of the following statement(s) is(are) correct? [Given : e^{-1}=0.36 ] [JEE Adv 2023 P1]

The value of the resistance $R$ is $3\;\Omega$ .
For $t < t_0$ , the value of current $I_1$ is $2\;\mathrm{A}$ .
At $t=t_0+7.2\;\mu\mathrm{s}$ , the current in the capacitor is $0.6\;\mathrm{A}$ .
For $t \rightarrow \infty$ , the charge on the capacitor is $12\;\mu\mathrm{C}$ .

Solution:

15 - I R = 6

I_1[3]=6

I_1=2\;\mathrm{A}

I=I_1+1=3

15-3R=6

\Rightarrow R=3\;\Omega

Eq. circuit is

\frac{1}{R_{\text{eq}}} = \frac{1}{3} + \frac{1}{3} + 1

R_{\text{eq}} = \frac{3}{5}\;\Omega

E_{\text{eq}} = 5+5+0=10

E_{\text{eq}} = 10 \times \frac{3}{5} = 6\;\mathrm{V}

Current in circuit is

\frac{6}{\left( \frac{3}{5} + 3 \right)} e^{-\frac{t}{C R}}

= \frac{6 \times 5}{18} e^{-\frac{7.2 \times 10^{-6}}{2 \times 10^{-6} \times 3 \cdot 6}}

= \frac{30}{18} \times e^{-1} = \frac{30}{18} \times 0.36

= 0.6\;\mathrm{A}

At steady state, voltage across capacitor

=6\;\mathrm{V}

Q=6 \times 2 = 12\;\mu\mathrm{C}

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