# Why is all of the energy from a battery stored on an inductor but only 50% on a capacitor?

I am learning about inductors and capacitors and we derived the energy stored on a capacitor to be 50% of that delivered by the battery. We did this considering a circuit of a capacitor connected to a battery and resistor in series, to not encounter the problem of an infinite initial current if we assumed there was no other resistor in the circuit. However our lecturer assured us that no matter how small the resistance in the circuit (even if it is just the small resistance of the wires), exactly 50% of energy would be lost. This made sense to me from the mathematics. I assume that in the case of a capacitor it is impossible to consider the theoretical case with no initial circuit resistance as you get infinities popping up in the mathematics?

Then we considered an inductor charging in a simple circuit consisting of just a battery and an inductor, and found that all of the energy from the battery is stored on the inductor. I appreciate that this is just a theoretical treatment and that some energy would be lost in the wires/internal resistance of the battery, and I also understand why a similar theoretical treatment of the capacitor case is impossible; however I can't think of the fundamental reason as to why it is completely impossible to charge a capacitor with anything but 50% of the battery energy whereas an inductor could theoretically store 100%.

• Note that one can charge a capacitor without losing energy with an ideal current source. The voltage source - inductor circuit is the dual of the current source - capacitor circuit. Similarly, if you try to charge an inductor with a current source, only 50% of the energy from current source will be stored in the inductor. – Alfred Centauri Nov 27 '16 at 23:38

When you try to force current through a superconducting inductor, the change of current will generate a back emf that will limit how much current can flow. The value of this back e.m.f. is $-L\frac{dI}{dt}$, and the work done by the current is the product of the current and the back emf. If the back emf is exactly equal to the voltage of the battery, current can flow (and can keep increasing - the rate of change of current is $\frac{dI}{dt}=-\frac{V}{L}$ ). This shows the current will increase linearly as all the energy of the power source is converted to magnetic energy - there is no need for a "loss" of energy in the transfer of energy from a battery to an inductor.

By contrast, when you start charging a capacitor, its initial voltage is zero. Electrons that start off with the full potential of the battery will have to lose most of that energy on their way to the capacitor, where they will only have a very small initial potential (since V=Q/C, and Q starts out at 0).

So in the inductor, the energy is actually stored in the B field; in the capacitor, it is stored in the electrons that came from the battery.

If you could "ramp" your battery (make its voltage increase as the capacitor is charging) you would be able to get (close to) 100% of the energy of the battery transferred. There are certain switching power supplies that try to mimic this type of thing by rapidly opening and closing a switch between source and load, with an inductor in series to smooth some of the power fluctuations that this would otherwise bring about.

• Beautiful explanation. I would add: it is just as valid to state the opposite view namely that the energy $\frac {1}{2}LI^2$ of the inductor is in the current $I$ rather in the field $B$ over the space, and similarly the energy of a capacitor resides between the plates in the field or in the electrons that are on the plates. – hyportnex Nov 27 '16 at 23:35
• The second sentence is, I'm afraid, false. For the zero resistance case as is stipulated by the OP, the inductor current is linear in time and thus, the emf is constant. – Alfred Centauri Nov 27 '16 at 23:41
• @AlfredCentauri fair point - sloppy phrasing. It is the change in current that drives emf. For a superconducting system, current will change linearly until the magnet quenches or you turn off the power supply... – Floris Nov 27 '16 at 23:45
• FWIW, the downvote isn't mine. – Alfred Centauri Nov 28 '16 at 0:20
• @AlfredCentauri thanks. Would you please see if you like the rewritten first paragraph better? – Floris Nov 28 '16 at 0:29

however I can't think of the fundamental reason as to why it is completely impossible to charge a capacitor with anything but 50% of the battery energy whereas an inductor could theoretically store 100%.

Essentially, to charge a capacitor with finite current from a voltage source requires dissipation. Why? Consider the ideal capacitor equation (in circuit theory):

$$i_C = C\frac{dv_C}{dt}$$

For $i_C$ finite, $v_C$ must be continuous. Since an uncharged capacitor has zero volts across, connecting an uncharged capacitor across a (non-zero) voltage source implies that, for finite current, some 'resistance' across which the voltage difference can drop. It follows that the energy stored in the capacitor is less than the energy delivered by the source since some energy is dissipated by the resistance (in whatever form it takes).

However, the ideal inductor equation is:

$$v_L = L \frac{di_L}{dt}$$

Note the difference here; specifically note that the voltage across can be discontinuous without implying infinite current. This is the crucial difference; the voltage across and current through are finite even for the ideal zero resistance case and so we can (ideally) charge an inductor with a voltage source without dissipation.