Question

JEE MAINS PHYSICS QUESTION SEPTEMBER 2020

Answer 1

A capacitor of capacitance C₀ is charged to potential V₀. now it is connected to another uncharged capacitor of capacitance C₀/2.

we have to find the heat loss in this process.

solution : This can be solved easily with help of formula, ∆U = -1/2 C₁C₂/(C₁ + C₂) (V₁ - V₂)²

here C₁ = C₀, C₂ = C₀/2 , V₁ = V₀ and V₂ = 0

so, heat loss = -1/2 [C₀ C₀/2/(C₀ + C₀/2)] × (V₀ - 0)²

= -1/2 [C₀/3 ] V₀²

= -1/6 C₀V₀²

Therefore the magnitude of heat loss in this process is 1/6 C₀V₀² i.e., option (3) is correct choice