Exams › NEET › Physics
A capacitor of 2μF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is:
- 0%
- 20%
- 75%
- 80%
Correct answer: 75%
Solution
When the switch is turned to position 2, the capacitor discharges through the circuit. The energy dissipated is given by the fraction of the initial energy lost, which is 75% in this case.
Related NEET Physics questions
- A positively charged rod is brought near an uncharged conductor. If the rod is then suddenly withdrawn, the charge left on the conductor will be :
- Two metal spheres, one of radius R and the other of radius 2R respectively have the same surface charge density σ. They are brought in contact and separated. What will be the new surface charge densities on them?
- In a region, the potential is represented by V(x, y, z) = 6x − 8xy − 8y + 6yz, where V is in volts and x, y, z are in meters. The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is:
- If potential (in volts) in a region is expressed as V(x, y, z) = 6xy − y + 2yz, the electric field (in N/C) at point (1, 1, 0) is:
- What is the electrostatic force between the metal plates of a capacitor?
- When a proton is accelerated through \( 1 \, \text{V} \), then its kinetic energy will be
⚔️ Practice NEET Physics free + battle 1v1 →