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A capacitor of capacitance \mathrm{C} has \mathrm{Q}. Its connected to an identical capacitor through a resistance. Find heat produced in the resistance.
 

Option: 1

\mathrm{\frac{Q^2}{C}}

 


Option: 2

\mathrm{\frac{Q^2}{2 c}}
 


Option: 3

\mathrm{\frac{Q^2}{4 c}}
 


Option: 4

\mathrm{\frac{Q^2}{3c}}


Answers (1)

best_answer

As the capacitors are identical, so final charge on each one is \mathrm{\frac{Q}{2}}. Therefore heat produced.

\mathrm{=E_i-E_f =\frac{Q^2}{2 c}-2\left[\frac{\left(\frac{Q}{2}\right)^2}{2 c}\right] }

\mathrm{=\frac{Q^2}{2 c}-\frac{Q^2}{4 c}=\frac{Q^{2}}{4c} }

Hence option 3 is correct.
 

Posted by

Riya

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