Question

A 12μF capacitor is charged by a 100V supply. The supply is then disconnected, and the charged capacitor is connected to another uncharged 3μF capacitor. How much electrostatic energy of the first capacitor is lost in the process of attaining the steady situation?

Answer 1

Explanation:

C₁ = 4µF, V₁ = 200V, C₂ = 2μF, V₂ = 0 - so, common potential difference across the two capacitors after connection is V: = C₁V₁ + C₂V₂ C₁ + C₂

4 x 10-6 × 200 +0 (4+2) × 10-6

=

133.33V

initially, total energy stored in capacitors before connection is U₂ = C₁V₁² - so, U₁ = x 4 x 10-6 × 200² = 0.08J

and total energy stored in capacitors after connection is Uf = (C₁+C₂)V² so, Uf = (4 + 2) × 10-6 × 133.33² : 0.053.J

so, energy lost due to connection is AU = Uf - U₂ AU = 0.053 -0.08 = -0.027 J