Charging the capacitor by using potential energy source or voltage source through a frictional media such as the electrical resistor will cause losing the electrical energy into friction, since the current passing to charge the capacitor will meet friction in the resistance. So, in the charging process the potential energy of the charging source is converted into kinetic energy where a fraction of it, commonly one half will be converted int heat and the other half will be stored.

In order to avoid this loss one first store the kinetic energy into an inductor and then transfer this stored energy into a potential energy in a capacitor to be charged.

So charging capacitor through kinetic energy storing element will be not lossy if the inductor is ideal.

That is substituting the lossy resistor by storing lossless inductor will resolvethe problem of the losses.

It is the same idea of the pendulum.

Best wishes

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html

Thanks Sir

Is there any intuitive explanation of  this question, because i am not able to clearly understand the answer from the above link ?

If you lower the resistance, the charge current will be higher. This means that the charge time will be shorter but also that the dissipation in the resistor be higher. In the end, the total energy loss is the same, independent of the resistor value. This effect only occurs if you charge a capacitor from a constant voltage source. Initially, the voltage drop over the resistor is equal to the source voltage, therefore the losses are high.

If you could charge the capacitor with a constant current source, the losses in the resistor can be much lower and ideally equal to zero. In practice this means that the source voltage has to go up together with the capacitor voltage. In this case the voltage drop over the resistor would be constant (because the current is constant) and therefore the dissipation in the resistor can be as low as you want.

This is an old problem that puzzles many ...Usual explanation is 1/2 energy loss is in heat (i^2R) or radiated if R approaches zero. This is indeed not very appealing although mathematically it is possible to show this using limits. A similar classic problem is paralleling a charged capacitor C with another uncharged  capacitor C when we lose 1/2 the energy using charge conservation principle.

Consider, if it helps, that the capacitor is like spring, the energy stored in a linear spring is1/2 kX^2. No matter how fast the spring is compressed the stored energy is dependent only on the final state X ( and of course constant k). even if we think of an impulsive force (dirac delta  ). Similarly however  fast  the capacitor is charged (ie R is reduced ) the final stored energy on the capacitor =1/2 C V0^2. Some say there is a connection to Thermodynamical principles.. but the phenomena is  counter intuitive..

Hope it helps..

Cheers

Discharging or charging a capacitor  intuitively you can see that the lower the resistance, the lower the discharge time (CxR), but current increases and the dissipation is porportional to the square of current, so it remains the same: 

E = k * I^2 * R * t = k * V^2 / R * t. = k * V^2 / R * C * R  = k * V^2 * C; independent of R,

k is the constant to taken in account because this is an exponential pulse, not a squarted one, and by integration it happens to be 1/2. 

If you charge a capacitor through a constant current source, from a constant voltage source, the dissipation in the current source is exactly the same. Supposing that the voltage of the capacitor reaches the voltage of the current source:

E= 1/2 * V * I * t  where t is the charging time.  (1/2 because the is a linear decrement of the voltage in the current source from V to 0).

Charging time t = C*V / I

E = 1/2 * V * I * C * V / I = 1/2 * V^2 * C

What happens if a charged capacitor is discharged over  a discharged one through a 0 resistance, i.e. using superconductors? Because the inductance of the circuit is not 0, there will be osicllations and the energy will be dissipated as electromagnetic waves.

Indeed Francesc, you are correct and my statement was somewhat poorly formulated. If the capacitor is charged from a constant voltage source, using a fixed or variable current (e.g. by using a variable resistor), the losses will always be the same, like you say. If the source voltage tracks the capacitor voltage, that is: if it is adjusted such that the charge current through a fixed resistor is constant, then the losses can be much lower.

Dear Jitender Mor,

Replace the resistor with an ideal inductor and an ideal diode. You will get different result for the energy in the capacitor after a period of oscillation of the LC tank. The reason is that the “kinetic” energy of the current of the inductor will convert in “potential” energy of the voltage across the capacitor.

 If the connection is by a resistor, as in your question and answers from the audience, then the “kinetic” energy in the dissipating element (the resistor) will transform in non-electrical energy (usually heat, but not only). Thus, the converted (dissipated) energy will not appear in the capacitor electrical energy. The math gives you the rates of dissipations and conversions. Half of the energy is dissipated in the resistor during the charging of the capacitor. The other half will be dissipated when you discharge the capacitor on a loading resistor.

The answers to your question depend on whether you have dissipating connection between energy source and energy storage, or you have a system of energy conserving elements. The electrical (and magnetic) energy conserving elements are ideal capacitors and ideal inductors (and ideal diode). None of those are available for purchasing. So, energy conversions and transmissions are always accompanied with energy losses, either in heat or other (e.g., electromagnetic) radiation. However, the sum of all energies remains the same, if you can “isolate” the system in an energy “box” (not possible either in practice, at least due to blackbody radiation). Here comes the problem with the energy source, in particular with the energy conservation principle (because the energy source pushes energy in the “box” from “outside of the box”), but this is beyond your question.

Sincerely,

Dr. O. Marinov

PS. I love the teachers’ quizzes. Watch their answers. The hyperphysics web site is a good resource for these quizzes.

Dear Simone Orcioni

Can there be an explanation to why a battery should be different, i.e. less energy lost when charging/discharging or would we loose at least 50% as in the capacitor case. see: https://www.researchgate.net/post/How_much_energy_is_lost_when_charging_a_battery

Best?

The apparent loss lies in the over simplified formulation that leads to the result (Q^2)/2C. Ideally, the energy stored in a capacitor at full charge is exactly (Q^2)/C, where is the charge stored.

When charging a capacitor by a battery abruptly ( one step), then always we have one half of the energy given by the battery is lost, which is the price that we have to pay to transfer the system from non-equilibrium state to an equilibrium one. The question where this energy loss go when the resistance goes to zero? This energy is transmitted to the connecting wires and the way this energy will be consumed depends on the availability of dissipation mechanism. however, this energy loss can be dramatically reduced in case of adiabatic charging.

Charging the capacitor by using potential energy source or voltage source through a frictional media such as the electrical resistor will cause losing the electrical energy into friction, since the current passing to charge the capacitor will meet friction in the resistance. So, in the charging process the potential energy of the charging source is converted into kinetic energy where a fraction of it, commonly one half will be converted int heat and the other half will be stored.

In order to avoid this loss one first store the kinetic energy into an inductor and then transfer this stored energy into a potential energy in a capacitor to be charged.

So charging capacitor through kinetic energy storing element will be not lossy if the inductor is ideal.

That is substituting the lossy resistor by storing lossless inductor will resolvethe problem of the losses.

It is the same idea of the pendulum.

Best wishes