Intially, switch S is connected to position 1 | vedclass

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Question

$U_{i}=\frac{1}{2} C \varepsilon_{1}^{2}$

$U_{f}=\frac{1}{2} C \varepsilon_{2}^{2}, \Delta U=\frac{1}{2} C\left(\varepsilon_{2}^{2}-\varepsilon_{1}

Vedclass · Accepted Answer

$\frac{C}{2}{\left( {{\varepsilon _1} + {\varepsilon _2}} \right)^2}$

Vedclass · Answer

$U_{i}=\frac{1}{2} C \varepsilon_{1}^{2}$

$U_{f}=\frac{1}{2} C \varepsilon_{2}^{2}, \Delta U=\frac{1}{2} C\left(\varepsilon_{2}^{2}-\varepsilon_{1}^{2}\right)$

$Q_{i n}=+C \varepsilon_{1}, Q_{f}=-C \varepsilon_{2}, \Delta Q=\left|Q_{j}-Q_{i}\right|$

$=C\left(\varepsilon_{2}+\varepsilon_{1}\right)$

Work done-by battery $W_{b}=\varepsilon_{2} \Delta Q=C\left(\varepsilon_{2}+\varepsilon_{1}\right) \varepsilon_{2}$

Heat generated $=W_{b}-\Delta U$

$=C \varepsilon_{2}^{2}+C \varepsilon_{1} \varepsilon_{2}-\frac{1}{2} C\left(\varepsilon_{2}^{2}-\varepsilon_{1}^{2}\right)$

$=\frac{1}{2} C\left(\varepsilon_{2}^{2}+\varepsilon_{1}^{2}+2 \varepsilon_{1} \varepsilon_{2}\right)=\frac{1}{2} C\left(\varepsilon_{1}+\varepsilon_{2}\right)^{2}$

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

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