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A capacitor of capacitance C is connected in series with a resistance R and a DC source of emf E through a key. The capacitor starts charging when the key is closed. By the time the capacitor has been fully charged, what amount of energy is dissipated in the resistance R? |
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Answer» `1/2 CE^2` (I )itspotentialenergystoredin thecapacitor (ii )heatenergydissipatedacrossresistance(R ) Nowwhenthecapacitorisfullychargedthenenergysuppliedby dcsource ` Q_(NET) . V=1/2CV^2+H ` `implies(CE )/(E )= 1/2CE^2+ H[ :.V = E ]` `impliesCE^2= 1/2CE^2+H impliesH= CE^2-1/2CE^2= 1/2 CE^2` |
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