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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. |
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Answer» Correct Answer - D 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` |
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