In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q_1 μ C . Then S is switched to position Q . After a long time, the charge on the capacitor is q_2 μ C . The magnitude of q_1 is _______.

Correct answer: 1.33. When S is connected to P, In steady state, i=(2−1)/(2+1) ⇒i=1/3 A Now, Charge on capacitor, q=CV ⇒q=1μF4/3 V ⇒q=4/3μC ⇒q=1.33 μC

Q45 / 57 · JEE Advanced 2021 (Paper 1) · Physics

In the circuit shown below, the switch $\mathrm{S}$ is connected to position $\mathrm{P}$ for a long time so that the charge on the capacitor becomes $q_{1} \mu \mathrm{C}$. Then $\mathrm{S}$ is switched to position $\mathrm{Q}$. After a long time, the charge on the capacitor is $q_{2} \mu \mathrm{C}$.

The magnitude of $q_{1}$ is _______.

Correct answer: 1.33

Solution

When $S$ is connected to $P$ ,
In steady state,

$i = \frac{2 - 1}{2 + 1}$

$\Rightarrow i = \frac{1}{3} A$

Now,

Charge on capacitor,

$q = C V$

$\Rightarrow q = (1 μF) (\frac{4}{3} V)$

$\Rightarrow q = \frac{4}{3} μC$

$\Rightarrow q = 1.33 μC$

Hence, the correct answer is 1.33.

Solution

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