Verified Solution • 2025–26 Session moderate

QJEEAdv23Phdeaf: JEE Advanced 2023 Physics - Class 12 JEEAdv23Phdeaf Solution: In a circuit shown in the figure, the capacitor is initially…

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 \mathrm{~A}

flows through the

1 \Omega

resistor. The key is closed at time

t=t_0

. Which of the following statement(s) is(are) correct?

[Given:

e^{-1}=0.36

]

JEE Advanced 2023 Paper 1 Online Physics - Capacitor Question 3 English

Multiple Choice Options

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

Step-by-Step Verified Solution

At

\mathbf{t}&lt;\mathbf{t}_0

,

JEE Advanced 2023 Paper 1 Online Physics - Capacitor Question 3 English Explanation 1

JEE Advanced 2023 Paper 1 Online Physics - Capacitor Question 3 English Explanation 2

JEE Advanced 2023 Paper 1 Online Physics - Capacitor Question 3 English Explanation 3

At

\mathbf{t} &lt; \mathbf{t}_0

\begin{aligned}i &amp; =\frac{\left(\frac{45}{3+R}\right)-5}{1=\frac{3 R}{3+R}} \\\\1 &amp; =\frac{45-15-5 R}{3+4 R} \\\\\Rightarrow R &amp; =3 \Omega\end{aligned}

(A) is correct.

I_1=\frac{\frac{15}{2}}{3+\frac{3}{4}} \Rightarrow I_1=2 \mathrm{~A}

(B) is correct.

At

\mathbf{t} &gt; \mathbf{t}_0

JEE Advanced 2023 Paper 1 Online Physics - Capacitor Question 3 English Explanation 4

\begin{aligned} \frac{\varepsilon}{r} &amp; =\frac{15}{3}+\frac{5}{1}+\frac{0}{3} \\\\ \frac{1}{r} &amp; =\frac{1}{3}+\frac{1}{1}+\frac{1}{3}=\frac{5}{3} \\\\ \varepsilon &amp; =10 \times \frac{3}{5} \Rightarrow 6 \mathrm{~V} \\\\ \text { At } t \rightarrow \infty &amp; \\\\ q &amp; =\mathrm{C} \varepsilon \\\\ &amp; =2 \times 10^{-6} \times 6=12 \mu \mathrm{C}\end{aligned}

\begin{aligned} &amp; \text { Time constant } \tau=R C \\\\ &amp; =(r+3) 2 \mu \mathrm{C} \\\\ &amp; =\left(\frac{3}{5}+3\right) \times 12=7.2 \mu \mathrm{s} \\\\ &amp; i=i . e^{-t / \tau} \\\\ &amp; i_0=\frac{\varepsilon}{r+3}=\frac{6}{\left(3+\frac{3}{5}\right)}\end{aligned}

\begin{aligned}\Rightarrow \quad i &amp; =\frac{6}{\frac{18}{5}} \times e^{-7.2 / 1.2} \\\\&amp; =\frac{5}{3} \times 0.36=0.6 \mathrm{~A}\end{aligned}

All options are correct.

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