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  • Initially switch S is connected to position 1 for a long time. The net amount of heat generated  in the circuit after it is shifted to position 2 is <img alt="" src="https://cdn.entr

Initially switch S is connected to position 1 for a long time. The net amount of heat generated  in the circuit after it is shifted to position 2 is

Option: 1

\frac{C}{2}\left ( \epsilon_{1} + \epsilon_{2} \right )\epsilon_{2}


Option: 2

C\left ( \epsilon_{1} + \epsilon_{2} \right )\epsilon_{2}


Option: 3

\frac{C}{2}\left ( \epsilon_{1} + \epsilon_{2} \right )^{2}


Option: 4

C\left ( \epsilon_{1} + \epsilon_{2} \right )^{2}


Answers (1)

best_answer

Energy Stored -

U=\frac{1}{2}CV^{2}=\frac{1}{2}QV=\frac{Q^{2}}{2C}

Ui=\frac{1}{2}C\epsilon _{1}^{2}    and  Uf=\frac{1}{2}C\epsilon _{2}^{2}     

So          \bigtriangleup U=\frac{1}{2}C\left (\epsilon _{2}^{2}-\epsilon _{1}^{2} \right )

Qin=+C\epsilon _{1}    ;       Qfinal=-C\epsilon _{2}  So  \Delta Q=-C\left (\epsilon _{1} +\epsilon _{2} \right )

work done by the battery  W_{b}=E_{2}.\Delta Q\mathrm {=C\left (\epsilon _{2} +\epsilon _{1} \right )\epsilon _{2}}

Heat generated = W_{b}-\Delta U=C\epsilon _{2}^{2}+C\epsilon _{1}\epsilon _{2}-\frac{1}{2}C\left (\epsilon _{2}^{2}-\epsilon _{1}^{2} \right )

=\frac{1}{2}C\left ( \epsilon _{2}^{2}+\epsilon _{1}^{2}+2\epsilon _{1}\epsilon _{2} \right )=\frac{1}{2}C\left (\epsilon _{1}+\epsilon _{2} \right )^{2}

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