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NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEEasy

Question

Two condensers, one of capacity CC and the other of capacity C/2C/2 are connected to a VV volt battery, as shown in the figure. The energy stored in the capacitors when both condensers are fully charged will be:

A

2CV22CV^2

B

14CV2\frac{1}{4}CV^2

C

34CV2\frac{3}{4}CV^2

D

12CV2\frac{1}{2}CV^2

Step-by-Step Solution

The total energy stored in a system of capacitors is the sum of the energies stored in individual capacitors. Based on the options, the capacitors are connected in parallel across the voltage VV.

  1. Energy in first capacitor (C1=CC_1 = C): U1=12C1V2=12CV2U_1 = \frac{1}{2} C_1 V^2 = \frac{1}{2} CV^2
  2. Energy in second capacitor (C2=C/2C_2 = C/2): U2=12C2V2=12(C2)V2=14CV2U_2 = \frac{1}{2} C_2 V^2 = \frac{1}{2} \left( \frac{C}{2} \right) V^2 = \frac{1}{4} CV^2
  3. Total Energy: Utotal=U1+U2=12CV2+14CV2=34CV2U_{\text{total}} = U_1 + U_2 = \frac{1}{2} CV^2 + \frac{1}{4} CV^2 = \frac{3}{4} CV^2

Alternatively, for parallel combination, Ceq=C+C/2=3C/2C_{\text{eq}} = C + C/2 = 3C/2. The total energy is U=12CeqV2=12(3C/2)V2=34CV2U = \frac{1}{2} C_{\text{eq}} V^2 = \frac{1}{2} (3C/2) V^2 = \frac{3}{4} CV^2 [NCERT Class 12, Physics Part 1, Sec 2.15, Eq 2.69].

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROSTATIC POTENTIAL AND CAPACITANCE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEcondenserscapacitycapacityconnectedbattery

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