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 combine system is

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{C}{4} {(V_{1} - V_{2})}_{1}^{2}$
Initial energy of the combined system
$U_{1} = \frac{1}{2} C V_{1}^{2} + \frac{1}{2} C V_{1}^{2} = \frac{C}{2} (V_{1}^{2} + V_{2}^{2})$
On joining the two condensers in parallel, common potential.
$V = \frac{C V_{1} + C V_{2}}{C + C} = \frac{V_{1} + V_{2}}{2}$
$∴$ Final energy of the combined system
$U_{2} = \frac{1}{2} (C + C) {(\frac{V_{1} + V_{2}}{2})}^{2}$
Decrease in energy
$Δ U = U_{1} - U_{2} = \frac{1}{2} C (V_{1}^{2} + V_{2}^{2}) - \frac{1}{2} (2 C) {(\frac{V_{1} + V_{2}}{2})}^{2}$
$= \frac{C}{2} [2 (V_{1}^{2} + V_{2}^{2}) - {(V_{1} + V_{2})}^{2}] = \frac{C}{4} [2 V_{1}^{2} + 2 V_{2}^{2} - V_{1}^{2} - V_{2}^{2} - 2 V_{1} + V_{2}] = \frac{C}{4} {(V_{1} - V_{2})}^{2}$

Suggest Corrections

Similar questions

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

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 capacitors are connected after charging. The decrease in energy of the combined system is :

Two capacitors of capacitance C and 2C, are charged to potential differences V ad 2Vrespectively. If the two positive plates are connected together and the two negative plates are connected together, then this system of capacitance

v_{1} + v_{2} = 4

and

v_{1}^{2} + v_{2}^{2} = 16

. Find value of

v_{1}

and

v_{2}

Q. A parallel plate capacitor of capacitance

C

is connected to a battery and is charged to a potential difference

V

.
Another capacitor of capacitance

2 C

is connected to another battery and is charged to potential difference

2 V

. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

Join BYJU'S Learning Program

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

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 combine system is