Two insulated metallic spheres of 3,mu F and | vedclass

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(C) The energy loss $\Delta U$ when two capacitors with capacitances $C_1$ and $C_2$ at potentials $V_1$ and $V_2$ are connected is given by the formu

Vedclass · Accepted Answer

$0.0375$

Vedclass · Answer

(C) The energy loss $\Delta U$ when two capacitors with capacitances $C_1$ and $C_2$ at potentials $V_1$ and $V_2$ are connected is given by the formula: $\Delta U = \frac{1}{2} \frac{C_1 C_2 (V_1 - V_2)^2}{C_1 + C_2}$.Given: $C_1 = 3\,\mu F = 3 	imes 10^{-6}\,F$,$C_2 = 5\,\mu F = 5 	imes 10^{-6}\,F$,$V_1 = 300\,V$,$V_2 = 500\,V$.Substituting the values:$\Delta U = \frac{1}{2} 	imes \frac{(3 	imes 10^{-6}) 	imes (5 	imes 10^{-6}) 	imes (500 - 300)^2}{(3 	imes 10^{-6} + 5 	imes 10^{-6})}$$\Delta U = \frac{1}{2} 	imes \frac{15 	imes 10^{-12} 	imes (200)^2}{8 	imes 10^{-6}}$$\Delta U = \frac{15 	imes 10^{-12} 	imes 40000}{2 	imes 8 	imes 10^{-6}}$$\Delta U = \frac{60 	imes 10^{-8}}{16 	imes 10^{-6}} = 3.75 	imes 10^{-2}\,J = 0.0375\,J$.

Two insulated metallic spheres of $3\,\mu F$ and $5\,\mu F$ capacitances are charged to $300\,V$ and $500\,V$ respectively. The energy loss,when they are connected by a wire is (in $,J$)

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