QJEEAdv17Phafbf: JEE Advanced 2017 Physics - Class 12 JEEAdv17Phafbf Solution: In Process 2, total energy dissipated across the resistance E D …

Question

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In Process 2, total energy dissipated across the resistance E_D is

Accepted Answer

Process 1 :

Voltage is set to V₀/3

Charge supplied =

{{C{V_0}} \over 3}

Energy supplied =

{{{V_0}} \over 3} \times {{C{V_0}} \over 3} = {{CV_0^2} \over 9}

Process 2 :

Voltage is raised to 2V₀/3

additional charge supplied =

{{2{V_0}C} \over 3} - {{{V_0}C} \over 3} = {{C{V_0}} \over 3}

Energy supplied =

{{2{V_0}} \over 3} \times {{C{V_0}} \over 3} = {{2CV_0^2} \over 9}

Process 3 :

Voltage is raised to V₀

additional charge supplied =

{V_0}C - {{2{V_0}C} \over 3} = {{C{V_0}} \over 3}

Energy supplied =

{V_0} \times {{C{V_0}} \over 3} = {{CV_0^2} \over 3}

Total energy supplied to circuit =

{{CV_0^2} \over 9} + {{2CV_0^2} \over 9} + {{CV_0^2} \over 3}

= {6 \over 9}CV_0^2 = {2 \over 3}CV_0^2

Final energy stored in capacitor =

{1 \over 2}CV_0^2

Therefore,

Energy dissipated E_D = Energy supplied

-

Energy stored

= {2 \over 3}CV_0^2 - {1 \over 2}CV_0^2

= {{4CV_0^2 - 3CV_0^2} \over 6} = {1 \over 6}CV_0^2

= {1 \over 3}\left( {{1 \over 2}CV_0^2} \right)

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