Energy charged capacitor

It is easily imaginable to be able to accumulate the electrical energy in the capacitor. However, this is also temporary. Because the insulation resistance of the capacitor is a value of not infinity but limited, energy is lost by a natural electrical discharge. The electrical energy cannot be saved for a long term like the battery in a word.

Calculation of energy charged capacitor with

Now, the voltage e [V] on the charged capacitor C [F] on the charge q [coulomb] Assume that is even saved.

q = C*e   or   e = q/C --- Expression (1)

A further small charge capacitor ⊿q [Coulomb] consider that charge.

⊿U = e*⊿q ---- Expression (2)

It becomes it. The energy charged the capacitor with U [J] is charge as for the expression (2) From 0 It obtains if it integrates into Q by q.

U = 0Q e*dq = 0Q (q/C)dq = Q2/2C --- Expression (3)

U [J] of the voltage capacitor charged E [V] with electrical energy of C[F] is as follows. From Q = CE

U = (1/2)CE2 --- Expression (4)

It becomes it.

Notes when it charges capacitor from DC power

If it charges the capacitor from the constant-voltage source, energy becomes half.

In this case, the energy of the capacitor becomes half of the energy supplied from DC power though might it charge the capacitor from DC power and this energy be used.

In a word, the energy supplied from the DC power of E [V] though the energy charged the capacitor with by the above-mentioned expression (4) is obtained Us [J] :.

Us = QE = CE2 ---- Expression (5)

Vicinity (4) type becomes half energy of the expression (5).

Where did this lost energy go?It has been consumed by the resistance of wiring. Even if this resistance is unlimitedly adjusted to 0, the half of energy is consumed by minute resistance. In this case, the charging current becomes infinity momentarily at the time of switched on.

Then, 0 does lose the resistance of the coil and wiring and if it is a method of the charging current's not becoming infinity, is not lost energy either?It thinks about the thing with which it charges with the circuit that connects the coil of the ideal between the power supply and the capacitor of the ideal now.

In this case, the capacitor resonates to the coil in parallel and the voltage of the capacitor comes and goes from 0 between twice the power-supply voltage indefinitely though the value of the current is limited. It settles down gradually actually in the power-supply voltage the value of the capacitor because there is minute resistance. The half of energy is still lost as for this.

Method of reducing a decrease in energy even if it charges capacitor

It exists in the capacitor, and it is an equipment using the electrical charge and discharge of the capacitor and the thing of the efficiency 50% or more will not exist in this world if energy is sure to become half if the electrical charge and discharge is repeated.

The method by which energy is not lost even if it charges the capacitor is to always charge it by almost the same voltage as the capacitor. Resistance is not used like turning on switching type DC-DC converter power supply and the capacitor is charged gradually. The voltage rises gradually when the movement of the DC-DC converter is seen with the oscilloscope though it is momentary.

There is a method by which energy is not lost even if it theoretically charges the capacitor. It is a method of the connection of the coil of the ideal to the constant-voltage source of the above-mentioned ideal and the charge to the capacitor.

When the voltage of the capacitor at this time increases to twice that of the power-supply voltage first (The current at this time is 0) If the circuit is separated, the capacitor can be charged without the loss. It is a unexamination whether this method is feasible.

It thinks about the circuit that charges with the capacitor when it stockpiles one's energy in the coil like the switching power supply of the fly backing converter type, and this energy is opened. When the energy of this coil was opened, it charged it if it connected it with the capacitor with the diode because the voltage was irregular by the same voltage as the capacitor.

Is it wrong anywhere though doesn't think that energy is theoretically lost if it is this method?

Application of energy accumulated in capacitor

What's useful and what is stored capacitor energy.