A capacitor of capacitance C is charged to voltage V then connected through a resistor R to an uncharged capacitor of same C. Final voltage across each:
AV/2
BV
C0
DV/4
Answer & Solution
Correct answer: A. V/2
**Charge conservation:** CV initial.
**After sharing equally:** each has CV/2 → V/2 voltage.
Energy lost as heat in R = (1/2)CV² - 2 × (1/2)C(V/2)² = CV²/4 (half lost regardless of R).
Keep practising Unit 2A: Newton's Laws and Translational Equilibrium
Browse all Unit 2A: Newton's Laws and Translational Equilibrium questions for AP Physics 1, or explore more Physics topics.
Related questions
At what temperature does kT equal 1 eV?A car travels 100 km at 50 km/h, then 100 km at 100 km/h. Average speed:Photon of wavelength 200 nm strikes metal with work function 2 eV. KE of fastest photoelecWork done in stretching wire by 1 mm is W. Work to stretch it by 2 mm (same wire, within eMagnetic flux through a coil changes from 5 mWb to 1 mWb in 0.1 s. Induced EMF:Bohr radius for hydrogen is 0.53 Å. Radius of 3rd orbit:Two coherent sources of intensity I and 4I interfere. Maximum intensity in resulting patteFor a body in SHM with amplitude A, at displacement x, ratio of KE to total energy: