Energy to charge capacitor and stored energy

In summary, the capacitor stores half the energy supplied by the battery. The power delivered by the battery in charging a capacitor of value 'C' to voltage 'V' is C*V^2. The energy stored in the capacitor is (C*V^2)/2. The difference energy i.e. (C*V^2)/2 is dissipated in the resistor. Even in case of a very large capacitor and very small resistor (as described by you), the initial current would be very high due to low resistor and would remain high for pretty long due to big capacitor which would mean high value of (I^2)*R. The current would exponentially decrease as the capacitor charges, with time constant RC
  • #1
sridhar10chitta
28
0
I use a 1V battery to charge a capacitor of capacity 1.602x10raised to -19 Farad (1.602x10raised to -19 coulomb is the electric charge of a single electron).
What is the energy spent by the battery and the energy stored in the capacitor ?
 
Physics news on Phys.org
  • #2
There's a formula for energy stored in capacitor which you'll need here. You can assume that there is no energy loss, ie. energy in capacitor = energy provided by battery.
 
  • #3
The capacitor stores half the energy supplied by the battery. Remaining half is lost in the resistance (resistance of wires used) as I*I*R. The energy lost in the resistance is independent of value or resistance and depends primarily on the capacitance.
 
  • #4
vinayakbhat82 said:
The capacitor stores half the energy supplied by the battery. Remaining half is lost in the resistance (resistance of wires used) as I*I*R. The energy lost in the resistance is independent of value or resistance and depends primarily on the capacitance.

This is not true. One could easily imagine a situation where you have a very large capacitor connected to a power source by a very resistive wire. The ideal case would have the resistance of the wire be zero whereby all the power received would be equal to the power delivered. But if half the power is always dissipated by the wire regardless of resistance, then this is a discontinuous relationship. I think you are thinking about the power delivered using an AC signal in a reactive circuit.

If he wants to account for the power lost due to ohmic losses then he can model the wire as a resistor in series with the capacitor and solve the new differential equation or use the appropriate transform to find the current relationship and then integrate over time to find the power delivered by the power source and the power dissipated by the resistor.
 
  • #5
Born2bwire said:
vinayakbhat82 said:
The capacitor stores half the energy supplied by the battery. Remaining half is lost in the resistance (resistance of wires used) as I*I*R. The energy lost in the resistance is independent of value or resistance and depends primarily on the capacitance.
This is not true.
I think it is true. However, the losses can be decreased by adding inductance. There is a good explanation here: http://www.smpstech.com/charge.htm
 
  • #6
Born2bwire said:
This is not true. One could easily imagine a situation where you have a very large capacitor connected to a power source by a very resistive wire. The ideal case would have the resistance of the wire be zero whereby all the power received would be equal to the power delivered. But if half the power is always dissipated by the wire regardless of resistance, then this is a discontinuous relationship. I think you are thinking about the power delivered using an AC signal in a reactive circuit.

If he wants to account for the power lost due to ohmic losses then he can model the wire as a resistor in series with the capacitor and solve the new differential equation or use the appropriate transform to find the current relationship and then integrate over time to find the power delivered by the power source and the power dissipated by the resistor.



As I understand, the statement made by me earlier is indeed true. Immaterial of values of capacitor and resistors, the capacitor would store half the energy supplied by the battery. The power delivered by the battery in charging a capacitor of value 'C' to voltage 'V' is C*V^2. The energy stored in the capacitor is (C*V^2)/2. The difference energy i.e. (C*V^2)/2 is dissipated in the resistor. Even in case of a very large capacitor and very small resistor (as described by you), the initial current would be very high due to low resistor and would remain high for pretty long due to big capacitor which would mean high value of (I^2)*R (Here I is instantaneous current integrated over time). The current would exponentially decrease as the capacitor charges, with time constant RC. Can you please clarify the "discontinuous relationship" being referred?
 
  • #7
TurtleMeister said:
I think it is true. However, the losses can be decreased by adding inductance. There is a good explanation here: http://www.smpstech.com/charge.htm

I'll have to work it out then by hand when I have some time then.

EDIT: Oh yeah it is, funny that.

[tex] P_{res} = IV = \int^\infty_0dt\frac{V_0^2}{R}\left(1-e^{-t/\tau_0}\right)e^{-t/\tau_0} \\
= \left.\frac{V_0^2}{R}\left[-\tau_0e^{-t/\tau_0} + \frac{\tau_0}{2}e^{-2t/\tau_0}\right]\right|_0^\infty \\
= \frac{V_0^2\tau_0}{2R} = \frac{V_0^2C}{2}[/tex]

Well then, you learn something new everyday.
 
Last edited:
  • #8
Thanks all of you. It is clear that there will be a loss in the wire...one electron or not...
Sridhar
 
  • #9
See, there is a basic difference between capacitor and cell. A cell supplies charge at constant potential, while the capacitor, transfers the charge at variable potentials, so in case of capacitor the formula is derived using integration.

while transfering charge q at potential V a cell does work qV, while in transfering charge q to a capacitor such that potential changes from 0 to V is 1/2 qV.
it is clear that while charging, half of energy is lost, at the same time, no charge is lost.
 
  • #10
TurtleMeister said:
I think it is true. However, the losses can be decreased by adding inductance. There is a good explanation here: http://www.smpstech.com/charge.htm

The energy loss is most likely present in the discharge phase of the capacitor, where the parasitic resistance in line with the load "steals" energy.

----------------
heh...after some simulations, in more complex circuits (like our beloved SMPSs) , the resistance present when charging a capacitor plays a big role in the efficiency of the circuit.
 
Last edited:
  • #11
If the resistance of wire and all is zero. Then will the energy supplied by battery = energy gained by capacitor = 1/2 CV^2 ??
 
  • #12
The energy supplied by the battery is Q x V.
The energy stored on the capacitor is 0.5Q x V
Energy can be lost in 3ways
1) resistance of the wire (this can be made =0)
2) any sparking at the switch(this can be made =0)
3) electro magnetic radiation from the connecting wires as the current changes during the charging (or discharging) (This cannot be made = 0)
Whenever coronet changes electro magnetic radiation occurs. The radiation could be radio waves and can be detected using a radio... It is crackling interference
 

1. How is energy stored in a capacitor?

Energy is stored in a capacitor through the separation of electric charges on the two plates of the capacitor. When a voltage is applied, one plate becomes positively charged and the other becomes negatively charged, creating an electric field between them. This electric field stores energy.

2. How does the amount of energy stored in a capacitor relate to its capacitance and voltage?

The amount of energy stored in a capacitor is directly proportional to both its capacitance and voltage. This means that increasing either the capacitance or the voltage will result in a greater amount of energy being stored in the capacitor. The equation for calculating the energy stored in a capacitor is 1/2 * capacitance * voltage^2.

3. Can a capacitor hold an unlimited amount of energy?

No, a capacitor has a maximum limit to the amount of energy it can hold, which is determined by its capacitance and voltage. If the voltage exceeds the maximum limit, the capacitor can become damaged or even explode. It is important to choose a capacitor with the appropriate capacitance and voltage rating for the intended use.

4. How long can a capacitor hold its stored energy?

A capacitor can hold its stored energy for an indefinite amount of time, as long as it is not connected to a circuit that allows the energy to discharge. However, over time, a small amount of energy may leak out due to the natural resistance of the capacitor's materials.

5. How is the stored energy in a capacitor used?

The stored energy in a capacitor can be used in a variety of ways depending on the circuit it is connected to. For example, in electronic devices, capacitors can be used to power displays, store data, or provide a quick burst of energy for a motor to start. In larger systems, such as power grids, capacitors can be used to regulate voltage and provide backup power in case of a power outage.

Similar threads

  • Electromagnetism
Replies
10
Views
486
  • Electromagnetism
Replies
16
Views
447
Replies
9
Views
1K
Replies
21
Views
515
Replies
6
Views
1K
Replies
9
Views
379
Replies
3
Views
2K
  • Electromagnetism
Replies
3
Views
836
  • Electromagnetism
Replies
4
Views
1K
Replies
7
Views
1K
Back
Top