1
$\begingroup$

For a simple circuit with a battery supplying a voltage V to a capacitor, let us assume that the charge on the capacitor is Q. Now, the work done by the battery or the energy supplied is given by the relation:

W = QV

But the energy stored in the capacitor is given by:

U = (1/2)QV

The value of Q as well as that of V should be the same in both the equations.

Now my question is, where is the other half of the energy that the battery supplied?

$\endgroup$
  • $\begingroup$ It's dissipated in the wiring resistance between the capacitor and the battery and the internal resistance of the battery while the capacitor charges. $\endgroup$ – CuriousOne Jun 5 '15 at 5:58
4
$\begingroup$

At the moment the circuit is completed, the capacitor has zero voltage, while the supply has $V$. This voltage difference creates an electric field that accelerates charges. This acceleration sets up a current.

As the current flows, the capacitor charges until the voltage reaches $V$ as well. At this point there is no voltage difference. But the accelerated charges are still moving. So half the energy has gone into the capacitor and (discounting losses) half has gone into the current in the wire. The current will continue to flow, charging the capacitor above $V$ until the current stops. This is overshoot. Then since a potential difference exists, current will flow back the other way. The current and voltage oscillate for a period. This oscillation behavior in the circuit is ringing. Resistance in the circuit will eventually remove this extra energy, leaving only the charged capacitor.

This is very similar to suspending a ball from a spring and releasing it. It can be slowly lowered to the new equilibrium point, or it can be dropped and it will oscillate above and below the new equilibrium until frictional losses remove the extra energy.

$\endgroup$
2
$\begingroup$

Half of the energy is lost to the battery's internal resistance (or other resistances in the circuit).if you try to consider an ideal battery with 0 internal resistance, the notion of charging the capacitor breaks down.since the capacitor and the battery are connected by a (0 resistance) wire, their voltages are the same the instant they are connected, no current flows from the battery to the capacitor.there is no charging.

$\endgroup$
  • $\begingroup$ The first sentence is correct but the rest of the answer is not. In the ideal case, the voltage across the capacitor is a step function and the current through is a delta 'function'. $\endgroup$ – Alfred Centauri Jul 11 '15 at 13:26
  • $\begingroup$ The answer is incorrect. $\endgroup$ – my2cts May 15 '18 at 18:47
1
$\begingroup$

Let's consider a Conductor (ideal) connected between two points having potential difference V.

enter image description here

So, Energy dissipated by the conductor when Q Charge passes through the ideal conductor connected between a potential difference of V Volt must be equal to the Energy supplied by the battery to the conductor which is equal to QV.

(NOTE: In case of Resistance also, the same amount of energy is dissipated. The only difference is that in a Resistance the dissipated energy appears as heat but in this case of an ideal conductor, it appears in some other form (maybe as a spark or Electromagnetic radiation, which I'm not sure)


Now let's consider the case of Charging of a Capacitor through ideal Conductor. :

Charging Circuit

Now did you figure out where Energy must be dissipated????

Yess !!! Exactly.. (VC-VB) is a non-zero quantity and hence Potential difference exists across the ideal conductor connected between the points C and B.

Using above discussion :

Calculations


[ Note : The model above is an oversimplification of everything. For example, why only the right wire dissipates energy? Actually, energy is dissipated throughout the ideal conductor in form of sparks or Electromagnetic radiations. But if you need to know why energy is dissipated, it turns out that it's not required to have a knowledge of exactly how energy is dissipated. I only tried to point out that. Ignore this answer totally if you are interested in knowing how ]

$\endgroup$
  • 1
    $\begingroup$ Why are you telling the reader that this is the answer they may not have wanted after they've finished reading the whole thing?! $\endgroup$ – Anurag B. May 15 '18 at 14:42
  • $\begingroup$ I have to otherwise split the PS part into two. Otherwise the For example part will be meaningless without reading the answer $\endgroup$ – Madhuchhanda Mandal May 15 '18 at 15:05
0
$\begingroup$

As pointed out in some comment, electrons are being accelerated in the process of charge. This generates electromagnetic radiation. Try doing the calculation using Poynting vector, the way Maxwell defined electromagnetic energy.

$\endgroup$
  • $\begingroup$ This is wrong. The acceleration is a transient phenomenon at vet low frequency. No radiation is involved. $\endgroup$ – my2cts May 15 '18 at 18:49
0
$\begingroup$

Your mistake is in assuming that the capacitor has a voltage V equal to that of the battery at $t = 0$. It does not. So your equation $W = QV$ is simply wrong. Any real circuit, to which Kirchhoff's laws apply, will have a resistance. By Kirchoff's law:

$V_b = IR + Q/C$

Differentiate to obtain:

$\frac{dI}{dt} = -I/(CR)$

Solve that equation with appropriate initial condition and you will find that the current and charge on the capacitor (and hence the voltage across it) approach the equilibrium value exponentially.

$\endgroup$
0
$\begingroup$

The equation $W=VQ$ is indeed wrong. During charging the voltage over the capacitor is on average $V/2$, so the energy stored in the capacitor is $VQ/2=CQ^2/2$.

No energy disappeared, neither by dissipation nor by radiation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.