In the figure shown, after the switch ‘ | vedclass

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Question

Before switch is shifted:

Energy stored,

$\mathrm{U}_{1}=\frac{1}{2} \mathrm{CE}^{2}$

and charge stored,

$Q=C E$

After switch is shifte

Vedclass · Accepted Answer

$\frac{3}{8}\frac{{{Q^2}}}{C}$

Vedclass · Answer

Before switch is shifted:

Energy stored,

$\mathrm{U}_{1}=\frac{1}{2} \mathrm{CE}^{2}$

and charge stored,

$Q=C E$

After switch is shifted:

$V_{M}-V_{N}=\frac{Q}{4 C}=\frac{C E}{4 C}=\frac{E}{4}$

Energy stored, $U_{f}=\frac{1}{2}(4 \mathrm{C})\left(\frac{E}{4}\right)^{2}=\frac{1}{8} \mathrm{CE}^{2}$

Energy dissipated $=U_{i}-U_{f}=\frac{3}{8} C E^{2}$

In the figure shown, after the switch $‘S’$ is turned from position $‘A’$ to position $‘B’$, the energy dissipated in the circuit in terms of capacitance $‘C’$ and total charge $‘Q’$ is

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