Two identical capacitors, have the same capacitance$$C$$. One of them is charged to potential $${{V}_{1}}$$ and the other to $${{V}_{2}}$$ the negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

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TuteeHUB · Accepted Answer

$$\frac{1}{4}C{{({{V}_{1}}+{{V}_{2}})}^{2}}$$

TuteeHUB · Answer

$$\frac{1}{4}C({{V}^{2}}_{{{1}^{{}}}}-{{V}^{2}}_{2})$$

TuteeHUB · Answer

$$\frac{1}{4}C({{V}^{2}}_{1}+{{V}^{2}}_{2})$$

TuteeHUB · Answer

$$\frac{1}{4}C{{({{V}_{1}}-{{V}_{2}})}^{2}}$$

1.	Two identical capacitors, have the same capacitance\[C\]. One of them is charged to potential \[{{V}_{1}}\] and the other to \[{{V}_{2}}\] the negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is
A.	\[\frac{1}{4}C({{V}^{2}}_{{{1}^{{}}}}-{{V}^{2}}_{2})\]
B.	\[\frac{1}{4}C({{V}^{2}}_{1}+{{V}^{2}}_{2})\]
C.	\[\frac{1}{4}C{{({{V}_{1}}-{{V}_{2}})}^{2}}\]
D.	\[\frac{1}{4}C{{({{V}_{1}}+{{V}_{2}})}^{2}}\]
Answer» D. \[\frac{1}{4}C{{({{V}_{1}}+{{V}_{2}})}^{2}}\]

Two identical capacitors, have the same capacitance\[C\]. One of them is charged to potential \[{{V}_{1}}\] and the other to \[{{V}_{2}}\] the negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

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