SATHEE: Electrostatics Question 648

Electrostatics Question 648

Question: 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

Options:

A) $\frac{1}{4} C (V_{1^{}}^{2} - V_{2}^{2})$

B) $\frac{1}{4} C (V_{1}^{2} + V_{2}^{2})$

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

D) $\frac{1}{4} C {(V_{1} + V_{2})}^{2}$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Initial energy of the system $U_{i} = \frac{1}{2} C V_{1}^{2} + \frac{1}{2} C V_{2}^{2}$

When the capacitors are joined, common potential $V = \frac{C V_{1} + C V_{2}}{2 C} = \frac{V_{1} + V_{2}}{2}$

Final energy of the system $U_{f} = \frac{1}{2} (2 C) V^{2} = \frac{1}{2} 2 C {(\frac{V_{1} + V_{2}}{2})}^{2} = \frac{1}{4} C {(V_{1} + V_{2})}^{2}$

Decrease in energy = $U_{i} - U_{f} = \frac{1}{4} C {(V_{1} - V_{2})}^{2}$

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Electrostatics Question 648

Question: Two identical capacitors, have the same capacitanceC . One of them is charged to potential V1 and the other to V2 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

Options:

Answer:

Solution:

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Question: 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