Energy to charge capacitor and stored energy

Main Question or Discussion Point

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 ?
 

Answers and Replies

Defennder
Homework Helper
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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.
 
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.
 
Born2bwire
Science Advisor
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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.
 
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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
 
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?
 
Born2bwire
Science Advisor
Gold Member
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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.
 
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Thanks all of you. It is clear that there will be a loss in the wire....one electron or not....
Sridhar
 
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.
 
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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.
 
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If the resistance of wire and all is zero. Then will the energy supplied by battery = energy gained by capacitor = 1/2 CV^2 ??
 
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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
 

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