The energy stored in a capacitor is given by the formula E = 1/2 CV^2.

For the first capacitor, the energy E1 is E = 1/2 CV^2.

For the second capacitor, the energy E2 is E = 1/2 * 2C * (2V)^2 = 4E.

When the two capacitors are connected, the total energy is the sum of the energies of the two capacitors, which is E1 + E2 = E + 4E = 5E.

The final energy when the capacitors are connected is given by the formula E = 1/2 C'V'^2, where C' is the total capacitance and V' is the total potential.

The total capacitance C' is C + 2C = 3C (since capacitors in parallel have their capacitances added), and the total potential V' is V (since capacitors in parallel have the same potential across them).

So, the final energy E' is E' = 1/2 * 3C * V^2 = 1.5E.

The loss of energy is the initial total energy minus the final energy, which is 5E - 1.5E = 3.5E.

Therefore, the loss of energy is x3E, where x is 3.5/3 = 1.17 (approximately).

Question

The energy stored in a capacitor is given by the formula E = 1/2 CV^2.

For the first capacitor, the energy E1 is E = 1/2 CV^2.

For the second capacitor, the energy E2 is E = 1/2 * 2C * (2V)^2 = 4E.

When the two capacitors are connected, the total energy is the sum of the energies of the two capacitors, which is E1 + E2 = E + 4E = 5E.

The final energy when the capacitors are connected is given by the formula E = 1/2 C'V'^2, where C' is the total capacitance and V' is the total potential.

The total capacitance C' is C + 2C = 3C (since capacitors in parallel have their capacitances added), and the total potential V' is V (since capacitors in parallel have the same potential across them).

So, the final energy E' is E' = 1/2 * 3C * V^2 = 1.5E.

The loss of energy is the initial total energy minus the final energy, which is 5E - 1.5E = 3.5E.

Therefore, the loss of energy is x3E, where x is 3.5/3 = 1.17 (approximately).

Knowee AI · Accepted Answer

The energy stored in a capacitor is given by the formula E = 1/2 CV^2.

For the first capacitor, the energy E1 is E = 1/2 CV^2.

For the second capacitor, the energy E2 is E = 1/2 * 2C * (2V)^2 = 4E.

When the two capacitors are connected, the total energy is the sum of the energies of the two capacitors, which is E1 + E2 = E + 4E = 5E.

The final energy when the capacitors are connected is given by the formula E = 1/2 C'V'^2, where C' is the total capacitance and V' is the total potential.

The total capacitance C' is C + 2C = 3C (since capacitors in parallel have their capacitances added), and the total potential V' is V (since capacitors in parallel have the same potential across them).

So, the final energy E' is E' = 1/2 * 3C * V^2 = 1.5E.

The loss of energy is the initial total energy minus the final energy, which is 5E - 1.5E = 3.5E.

Therefore, the loss of energy is x3E, where x is 3.5/3 = 1.17 (approximately).

A capacitor of capacitance C and potential V has energy E. It is connected to another capacitor of capacitance 2C and potential 2 V. Then the loss of energy is x3E, where x is_____

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