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 2 is \!\! \!\! .

Physics · Capacitors and R-C Circuits

JEE Advanced 2021 — Paper 1 — Question 35

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

$\mathrm{q}_{1} \mu \mathrm{C}$ . Then S is switched to position Q . After a long time, the charge on the capacitor is

$\mathrm{q}_{2} \mu \mathrm{C}$ .

The magnitude of ${{\text{q}}_{2}}$ is $\text{ }\!\!~\!\!\text{ }$ .

Answer: 0.67

Numerical answer — enter this value.

Step-by-step solution

After long time i circuit $=\frac{2}{3}$ ampere ${{\text{v}}_{\text{AB}}}=2-2\left( \frac{2}{3} \right)=\frac{2}{3}$ volt So, ${{\text{q}}_{2}}=\text{CV}=\left( 1 \right)\left( \frac{2}{3} \right)=\frac{2}{3}\mu \text{C}$

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Exam: JEE Advanced 2021
Paper: Paper 1
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Charging and Discharging of R-C Circuits

Answer: 0.67

Step-by-step solution

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