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

Question

Two identical capacitors, have the same capacitance $C$ . One of them is charged to potential $V_{1}$ and the other $V_{2}$ . 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 [2002]

Smart Achievers · Accepted Answer

(3)

Energy of a capacitor $V_{1}$

Initially, $U_{i} = \frac{1}{2} C V_{1}^{2} + \frac{1}{2} C V_{2}^{2} = \frac{1}{2} C (V_{1}^{2} + V_{2}^{2})$

After contact, common potential $V = \frac{V_{1} + V_{2}}{2}$

Finally, $U_{f} = \frac{1}{2} (2 C) {(\frac{V_{1} + V_{2}}{2})}^{2} = \frac{C}{4} {(V_{1} + V_{2})}^{2}$

$∴$ $Δ U = \frac{1}{2} C V_{1}^{2} + \frac{1}{2} C V_{2}^{2} - \frac{C}{4} {(V_{1} + V_{2})}^{2}$

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

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

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

Solution

Q.
Two identical capacitors, have the same capacitance $C$ . One of them is charged to potential $V_{1}$ and the other $V_{2}$ . 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 [2002]

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

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

3 $\frac{1}{4} C {(V_{1} - V_{2})}^{2}$

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

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Solution

Q. Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other 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 [2002]

1 14C(V12-V22)

2 14C(V12+V22)

3 14C(V1-V2)2

4 14C(V1+V2)2