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# Capacitance transforms electrical energy into what other form?

I've been working on my understanding of electricity. So I am a little ignorant on the subject. From what I've read, there are 3 fundamental ways energy is consumed in a circuit: resistance (R), capacitance (C), and inductance (L). But that's just the terminology, right? Energy is never consumed, it's transformed. So resistance primarily creates heat. Inductance primarily creates magnetism or a magnetic field. Capacitance??? Capacitance transforms electrical energy into what other form of energy?

## Discussions

Best Answer 10 years ago

You said that for an inductor energy is stored in a magnetic field.

A capacitor stores energy in an

electric fieldbetween its plates.http://en.wikipedia.org/wiki/Electric_field#Energy_in_the_electric_field

I don't know if you actually believe in electric fields, or magnetic ones, but

according to legendit takes energy to establish either one of these, a electric field, or a magnetic one, even in empty space. It's sort of a weird idea, the idea that you can do work upon, and get work back from, empty space, but that's the story.The actual formulas for the energy density due to E and B in of some small volume in space are U

_{E}=(1/2)*ε*E^{2}and U_{B}=(1/2)*(1/μ)*B^{2}Both U

_{E}and U_{B}have units of energy density: joules per cubic meter in SI units. To findall the energystored in a capacitor or inductor you integrate over all the space surrounding it. When doing this there are certain geometries that make the math easier than others.The math for a capacitor with wide flat plates is easy. The E field is essentially constant and only between the plates.

The math for an infinitely long inductor, or a thin toroid, is easy. The B field is essentially constant and only inside the coil.

By the way, the values of ε and μ are different inside matter than they are in empty space, and this maybe reflects the fact electric, or magnetic, fields interacting with matter are storing energy in a different way. For example, for a capacitor with vacuum, or air, between the plates, you can say the

energy is being stored in the empty space itself. However for a capacitor with some kind of dielectric, the energy is instead being stored in some microscopic interaction with matter, e.g.stretching or twisting individual molecules, much like tiny little springs.Answer 10 years ago

I've got all the equations, and can do most of the calculations for things that have already been designed. But I still lack understanding about Resistance, Inductance and Capacitance. Check this video out. This blew my mind.

It's #11 "Conspirators or the Treason" in the movie section after you go to the link. It wouldn't let me copy the quicktime link.

http://www.tufts.edu/as/wright_center/personal_pages/bob_m/

Answer 10 years ago

The direct link for this movie is:

http://www.tufts.edu/as/wright_center/personal_pages/bob_m/movie11.html

or

http://www.tufts.edu/as/wright_center/personal_pages/bob_m/franklin_prank.mov

That looks like a neat trick, Ben Franklin shocking his friends with a framed portrait of King George, but I don't think it reveals much about R, L, and C.

In fact, instead of all these centuries old references you keep mentioning, I think you might have better luck with something like:

http://www.allaboutcircuits.com/

It's a tutorial on electronic circuits, including passive components like R, L, and C, plus fancy stuff like diodes and transistors and op-amps and whatnot. Not to say that the centuries old stuff is not meaningful, but the stuff written more recently, as a tutorial, might be easier to understand. Of course you can read, or ignore, whatever you want to, including too-long replies like this one.

Back to the subject of R, L, and C, I'm not sure exactly what answer you are looking for. Maybe you want to know why R, L, and C might be described as "fundamental"?

I'm going to

guessa reason for this is because the formulas used to define them are similar to one another, differing only by a time derivative.I

_{C}(t)= C*(dV_{C}(t)/dt)I

_{R}(t)= (1/R)*V_{R}(t)I

_{L}(t)= (1/L)*integral(V_{L}(t)*dt)In words:

The current flowing into a capacitor is proportional to the time derivative of the voltage across the capacitor.

The current flowing through a resistor is proportional to the voltage across the resistor.

The current flowing through an inductor is proportional to the time integral of the voltage across the inductor.

All the derivatives and integrals are kind of annoying, and one way to get rid of them is to assume that the voltage, across R or L or C, is sinusoidal with frequency ω

If V

_{(R,L,or C)}= e^{j*ω*t}, then:I

_{C}= (j*ω*C)*V_{C}I

_{R}= (1/R)*V_{R}I

_{L}= (1/(j*ω*L))*V_{L}There's definitely some sneaky mathematical slight-of-hand that took place there. The most troubling of which is that V and I actually have a different meaning now. In the first equations above they were V(t) and I(t), real valued functions of time. In contrast the V and I now "complex amplitude". There are essentially three numbers needed to describe a sinusoid. These are: frequency, amplitude, and phase. Complex amplitude contains both amplitude and phase.

What this all means is that if you apply a sinusoidal voltage to a resistor you'll get a current that is exactly in phase with the applied voltage. If you apply the same sinusoidal voltage to a capacitor, you get a current that's 90 degrees out of phase. Apply the same sinusoidal voltage to an inductor, and the current is minus 90 degrees out of phase.

A consequence of this is, among R, L and C, only the resistor dissipates any power in a time-averaged sense. That's because V

_{R}(t) and I_{R}(t) are perfectly in phase. What I mean by "time-averaged" power is the total energy absorbed or released during one cycle, divided by one cycle period. In contrast, an inductor or capacitor is absorbing energy exactly half the time, and releasing energy the other half of the time.And this is exactly what other answerers have said already:

inductors and capacitors do not consume energy, turning it into heat the way a resistor does. Instead theystoreenergy. The mechanism by which the energy is stored, I've already mentioned: inductors store energy in magnetic fields and capacitors store energy in electric fields.By the way, storing energy does not have to be mysterious. There are other examples of physical things that store energy and give it back. A metal spring can store energy. A moving mass can store energy. In fact, the equations describing springs and masses look similar to those describing R, L, and C.

F=m*(d2x/dt2)=m*(dv/dt) [force to accelerate mass]

F=b*(dx/dt)=b*v [force due to friction]

F=k*x=k*integral(v*dt) [force to compress spring]

I

_{C}= C*(dV_{C}/dt) [current through capacitor]I

_{R}= (1/R)*V_{R}[current through resistor]I

_{L}= (1/L)*integral(V_{L}*dt) [current through inductor]Anyway, I think that's all I've got to say about R, L, and C. Hopefully allaboutcircuits, or some other reference can explain the rest.

Answer 10 years ago

Thanks, I'll check out the site. :-)

Answer 10 years ago

What legend ? You can show it experimentally.

Answer 10 years ago

Presumably it can be shown experimentally. But there is a certain amount of folklore involved too. Mostly I'm just retelling a story, as it was told unto me by my physics professor, and as it was told unto him by his physics professor, going back as least as far as James Clerk Maxwell, 1861.

Answer 10 years ago

That's one of the reasons why I'm asking. I started going back and reading all the old papers to work on my understanding of everything electrical and magnetic. Heresy?...probably. :-) But their books and papers are fascinating. It's worth the time. From what I've read, it almost seems like capacitance creates some sort of pressure, at high voltages.

I finished Maxwell's 1864 papers and JJ Thompson's book "Electricity and Matter". But I'm still on chapter 3 of Steinmetz "Elementary Lectures on Electric Discharges, Waves, Impulses and other Transients". It's deep, and I'm not all that bright. I have so many questions about why they changed the old model.

Answer 10 years ago

You're not doing work on "empty space," nor getting work therefrom. You're doing work on

the field. Space isn't empty; it is filled withthe field(or fields, depending on how GUTsy you want to be).When you increase the number of quanta in the electromagnetic field, then you must be doing work (since you can't create energy from nothing), and when you decrease the number of quanta, then you must be getting work back (since you can't destroy energy into nothing).

Answer 10 years ago

Best Answer

10 years ago

Actually, in AC systems, energy is radiated away by changing currents.

10 years ago

Capacitance and inductance do not consume energy; ideally, they can return all the energy put into them.

Resistance does dissipate energy by turning it into heat. Real-world capacitors and inductors generally do have some resistance losses.

Answer 10 years ago

+1

10 years ago

Capacitance works off the electric field. It stores electrons, much (as some hate to hear) like a bucket stores water.

Inductance works off the magnetic field. It can be modeled as a flywheel from mechanics

A resistor is, in fluid terms, a restriction, like a pipe of finite diameter. It works by dissipating energy through heating.

10 years ago

I have never heard anyone talk about electronic components in that way. But you may be on to something. The best answer i can give you is "STORED" energy. Capacitors can store a DC voltage for a short amount of time. If the voltage is an ALTERNATING voltage, then it will conduct the frequency according to this equation.... Xc=2(pi)FC .. where Xc is capacitive reactance, 2 is 2, F is the frequency in Hz, ... and C is the capacitance in FARADS.

Answer 10 years ago

Agreed,

Resistors resist current by burning it off as heat

Inductors create magnetic fields which depending on the application (transformer, inductor) can emit energy.

Capacitors act like batteries essentially, sucking up any voltage above their current charge level. When their charge equals the circuit voltage, the capacitor is 'full' (for arguments sake) and the charge in the capacitor pushes back against the circuit exactly as hard as the circuit pushes on the cap -- no energy flows. None is lost (ideally). They do have a small internal resistance, so technically they do have a loss, but its resistive.