Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

Question

Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is (A) (C/4)(V12 - V22) (B) (C/4)(V12 + V22) (C) (C/4)(V1 - V2)2 (D) (C/4)(V1 + V2)2

Tardigrade Admin · Accepted Answer

Total energy of two identical capacitors before connection.

E_{1} = \frac{1}{2} C V_{1}^{2} + \frac{1}{2} C V_{2}^{2}

.....(i)
When negative end of one capacitor is connected to negative end of other capacitor and positive to positive, then such type of connection is known as parallel connection.
Hence, equivalent capacitance after connection is given by,

C_{e q} = C + C

C_{e q} = 2 C

Common potential,

V = \frac{T o t a l c ha r g e}{T o t a l c a p a c i t an ce}

= \frac{Q _{1} + Q _{2}}{C + C}

= \frac{C V _{1} + C V _{2}}{2 C}

= \frac{V _{1} + V _{2}}{2}

∴

Total energy after connections

E_{2} = \frac{1}{2} \cdot C_{e q} \cdot V^{2}

= \frac{1}{2} \cdot 2 C (\frac{V _{1} + V _{2}}{2})^{2}

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

∴

Decrease in energy

= E_{1} - E_{2}

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

= \frac{C}{4} [2 V_{1}^{2} + 2 V_{2}^{2} - (V_{1}^{2} + V_{2}^{2} + 2 V_{1} V_{2})]

= \frac{C}{4} [V_{1}^{2} + V_{2}^{2} - 2 V_{1} V_{2}] = \frac{C}{4} (V_{1} - V_{2})^{2}

Q. 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 capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

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

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

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

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

Q. Two identical capacitors have the same capacitance C. One of them is charged to potential V1​ and the other to V2​. The negative ends of capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

4C​(V12​−V22​)

4C​(V12​+V22​)

4C​(V1​−V2​)2

4C​(V1​+V2​)2

Q. 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 capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

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

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

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

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