99 Percentile Qs Bank for JEE MainPhysicsCapacitance
In the following figure C₁=5 F , C₂=C₃=10 F and =20 ~V . Initially the switch S is connected to point A until capacitor C₁ is fully charged. Afterwards switch is thrown to left side and connected to point B . The charge on capacitor C₃ after equilibrium is reached will be
Options
- A40 C
- B100 C
- C50 C
- D20 C
Correct answer
A. 40 C
Step-by-step solution
Given that, C₁=5 F , C₂=C₃=10 F and E=20 ~V When the switch is connected to point A . Then, situation of the circuit is shown below. Charge stored in C₁ , Q=C₁ V₁=C₁ E=5 20 C =100 C When switch is connect to point B , then situation circuit is shown below. The charge Q₁ will start to distribute through, C₂ and C₃ to make the potential difference V across all the three capacitors C₁, C₂ and C₃ . Now, common potential difference, V= Total charge Total capacity = C₁ V₁+C₂ V₂+C₃ V₃ C₁+C₂+C₃ where, V₁=20 ~V , but V₂=V₃=