A capacitor of capacitance `5muF` is connected to a source of constant emf of `200 V`,Then switch was shifted to contact 2 from contact 1. Find the amount of heat generated in the `400Omega` resistance.

Answer

D

Explanation

Potential energy stored in the capacitor
`U=1/2 CV^2=1/2xx5xx10^-6xx(200)^2=0.1J`
During discharging this `0.1J` will distribute in direct ratio of resistance
`:. H_400=400/(400+500)xx0.1`
`=44.4xx10^-3J`
`=44.4mJ`