JEE Advanced 2023 · Paper 1 · Physics Q3

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PhysicsMultiple correct+4 / -2Question 20 of 51

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 $1\,\Omega$ resistor. The key is closed at time $t = t_0$ . Which of the following statement(s) is(are) correct?

[Given: $e^{-1} = 0.36$ ]

Four branches connected in parallel between a left vertical wire and a right vertical wire: (1) a 15 V cell in series with resistor R; (2) a 5 V cell in series with 1 Ω; (3) a 3 Ω resistor carrying current I₁ (arrow pointing left); (4) key K, capacitor C = 2 μF and a 3 Ω resistor in series. Both cells have the same orientation (positive plate on the same side).

The value of the resistance

R

3\,\Omega

For

t < t_0

, the value of current

I_1

is 2 A.

t = t_0 + 7.2\ \mu

s, the current in the capacitor is 0.6 A.

For

t \to \infty

, the charge on the capacitor is

12\ \mu

Answer and worked solution

Official answer: A, B, C, D

The three branches with sources and resistors join the same two nodes. Let $V$ be the potential difference between the nodes before the key is closed.

Before closing ( $t<t_0$ ).

The 5 V branch carries 1 A through $1\,\Omega$ , so $V=5+1=6$ V.
Then $I_1=6/3=2$ A. (B) true.
The 15 V branch must supply $1+2=3$ A, so $15-3R=6$ and $R=3\,\Omega$ . (A) true.

After closing. Seen from the capacitor branch, the rest of the circuit is a Thevenin source:

$V_{th}=6$ V;
$R_{th}=3\parallel1\parallel3=0.6\,\Omega$ .

So $\tau=(3+0.6)(2\,\mu\text{F})=7.2\,\mu$ s and
$i(t)=\frac{6}{3.6}e^{-(t-t_0)/\tau}.$

At $t-t_0=\tau$ : $i=1.667\times0.36=0.6$ A. (C) true.

Final charge. $q_\infty=CV_{th}=2\times6=12\,\mu$ C. (D) true.

Topics:RC circuits Kirchhoff's laws

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