
#1
May1507, 03:18 PM

P: 43

A simple circuit consists of a battery(emf=V), a switch and a capacitor(C).
At the steady state, the energy stored in the capacitor is 0.5CV^2. However, the work done by the battery is QV=CV^2 (move a total charge of Q across a potential difference of V). If the energy dissipated in the wire is negligible, according to the law of conservation of energy the work done by the battery=the energy stored in the capacitor. But 0.5CV^2!=CV^2. What's wrong with my reasoning above? Thanks! BTW, assuming the battery is without inner resistence(r=0). PS. When an resistor is connected in series with the capacitor, things are much easier. the rest of the work done by the battery (0.5CV^2) dissipates in the resistor. 



#2
May1507, 03:38 PM

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Nothing's wrong with your reasoning. It just happens that a zeroresistance wire is unphysical, as is a capacitor with zero equivalent series resistance. Such a circuit would have an instantaneous current of infinite magnitude  a delta function  and the energy stored would be undefined.
 Warren 



#3
May1507, 04:20 PM

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P: 28,801

If you sum up the work done by the battery, you get that 0.5. Zz. 



#4
May1507, 04:23 PM

P: 2,265

Energy stored in a capacitor
the energy stored in a capacitor is
[tex] \frac{1}{2} C V^2 [/tex] if [itex] V_{max} [/itex] is the maximum voltage a capacitor can hold without a breakdown of the dielectric material between plates, then the maximum energy a capacitor can hold is [tex] \frac{1}{2} C V_{max}^2 [/tex] it can be shown that, for a given dielectric material (which can withstand a limited E field), the maximum energy that a capacitor can hold is proportional to its volume. that is both satisfying and intuitive, in my opinion. 



#5
May1507, 07:22 PM

P: 43

By integral, it's pretty easy to show that the work done by the battery on the capacitor is 0.5CV^2. But I was thinking about what happened inside the battery. Assuming the battery has a constant potential V, then moving a total charge of Q from the anode to the cathode inside the battery. Therefore, the "total" work done by the battery is QV=CV^2, but no energy is lost in the circuit(no resistance in the circuit), where does the extra 0.5CV^2 go? 



#6
May1507, 07:23 PM

P: 43





#7
May1507, 07:33 PM

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#8
May1507, 11:17 PM

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#9
May1607, 09:52 AM

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Yet another point. Assuming you can achieve zero resistance you will find however that there is still an impedance due to the self inductance of the wire leads. Even if this self inductance is infinitesimally small it will result in the capacitor actually overcharging to a voltage greater than that of the battery. Think of the inductance as an "effective inertia" in the flow of the current. The circuit will then oscillate at a frequency:
[tex]\omega= 2\pi f = \sqrt{C/L}[/tex] between the initial zero charge on the capacitor and a maximum greater than the voltage of the battery (I think twice but will have to work it out). If the circuit is not shielded the energy will radiate away as em waves and the voltage of the capacitor will be a decaying oscillation centered around the battery voltage. Eventually it will settle down to the static case and you'll find the missing energy is exactly the energy of the radiated waves. Note that the amount of energy dissipated is independent of the impedance value and so is the same in the limit of zero impedance. The only variable is the complex phase of the impedance (how much is real = resistive and how much is imaginary = inductive) and this tells you how the energy is dissipated, either as heat or as EM radiation. If you prevent both from occurring then you will never get the assumed rest state. Here is a mechanical analogue. Begin with a spring at rest. Calculate the displacement of the spring for a certain force. Now connect a piston to the spring and apply a pressure equal to that force. Assume the piston and hydraulic fluid are frictionless. Open up the valve and you will find that the piston doesn't just shoot to the displacement you calculate and stop but rather will overshoot and you have an oscillating mass spring system. If you simply calculate the work due to the hydraulic force applied to the piston and the energy stored in the spring you will get the same type of mismatch. In this case you are neglecting the kinetic energy of the pistons motion. Another way to see your error is to recognize that as the capacitor charges its voltage must increase in proportion to the charge while as the battery discharges its voltage is (in the ideal case) constant. Since the work done is the integral of the voltage times dQ you must get different answers due to the fact that charge is conserved. The deficit is then obvious since there is an obvious difference in voltage between where the wire connects to the battery and where the wire connects to the cap. This is due to the impedance of the wire which accounts for both resistance and inductance. This is to say work must be done on the wire either as heat or the induction of a magnetic field around the wire. You must be careful in using DC assumptions to analyze transient phenomena. It would only be appropriate to think in DC terms when calculating steadystate power dissipation. Once you eliminate time independence you must always take into account those AC quantities especially in this case inductive impedance. Regards, James Baugh 



#10
Dec1307, 07:17 PM

P: 27

I read with keen interest your replies regarding the process of energy transfer when charging a capacitor to the question posed by Andy_ToK.
Quote "However, the work done by the battery is QV=CV^2 (move a total charge of Q across a potential difference of V).If the energy dissipated in the wire is negligible, according to the law of conservation of energy the work done by the battery=the energy stored in the capacitor. But 0.5CV^2!=CV^2. What's wrong with my reasoning above? Thanks! BTW, assuming the battery is without inner resistence(r=0)." Unquote I am confused as to whether this energy is taken into account or missing in the following equation with reference to AC power: For a perfect capacitor or inductor on the other hand there is no net power transfer, so all power is reactive. Therefore for a perfect capacitor or inductor: P = 0; modulusQ = S = Vrms.Irms = Irms^2 modulusX = Vrms^2/modulusX Where X is the reactance of the capacitor or inductor. (I have copied the formula and text from http://en.wikipedia.org/wiki/AC_power) Will be happy if you could clarify. Regards Sridhar 



#11
Dec1407, 09:34 PM

P: 997

The energy stored in a capacitor is indeed 0.5*C*V^2. In addition, when charging this capacitor from a voltage source, another 0.5*C*V^2 worth of energy is lost in the resistance as dissipation. Hence if we charge and fully discharge the cap at some frequency f, then the total power is given by C*V^2*f.
When charging a cap from a *voltage* source, the amount of energy lost equals the amount transferred to the cap. Inductors and/or current sources can charge a cap with much less loss. I hope this helps. Claude 



#12
Dec1507, 09:38 AM

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#13
Dec1507, 09:43 AM

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#14
Dec1507, 10:20 AM

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P: 16,477

You could probably also do it by making a RC circuit and taking the limit as R goes to zero. I am just assuming, but I think that limit would also be the expected result. 



#15
Dec1507, 04:06 PM

P: 167

Andy_ToK:
So Another thing that is not possile is that the wires have no inductance. In this case I dont know the answer. ( In the analysis of Switched Capacitor Circuits, as I studied, you dont take into account resistance ) 



#16
Dec2107, 12:43 AM

P: 27

Qupte:
I am confused as to whether this energy is taken into account or missing in the following equation with reference to AC power: For a perfect capacitor or inductor on the other hand there is no net power transfer, so all power is reactive. Therefore for a perfect capacitor or inductor: P = 0; modulusQ = S = Vrms.Irms = Irms^2 modulusX = Vrms^2/modulusX Where X is the reactance of the capacitor or inductor. (I have copied the formula and text from http://en.wikipedia.org/wiki/AC_power) Unquote Here is a plausible answer: In the AC case we are talking about steady state. Even if there were losses during an initial turnon transition, it would die out in a couple of cycles. The formula shown here applies to the situation *after* that transient behavior. Also, the equations assume that the driving source is in fact capable of absorbing energy during the half cycle when power is returned back to the source. So the observations are correct. These formula seem to hide the basic inevitability of transient losses that should occur when the system is just turned on. 



#17
Dec2107, 09:28 AM

P: 121

cant we just say that energy that is lost is in resistor only, as the wire has resistance anyways. or to say more, we can also say that the loss in energy is shared by the inductance and resistance that is inherited in the wire.
pls clarify 



#18
Dec2307, 08:51 AM

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PF Gold
P: 1,767

Try to assume away all the possible forms of loss and you must end up with a bouncing rock. Regards, James Baugh 


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