Good questions, and glad to see you're willing to ask them! It's a great topic, and really benefits one to know it beyond just a formula or two. Here are some links to start. Not great IMHO, but a start. Most treat it from a freq. or AC point of view, which is the most common, but I think you're asking about a single period of time, ie, t=0 to t=t_saturate. Understanding that is the best part! http://williamson-labs.com/480_rlc-l.htm http://www.allaboutcircuits.com/vol_1/chpt_15/1.html http://www.allaboutcircuits.com/vol_2/chpt_3/2.html http://www.play-hookey.com/dc_theory/rl_circuits.html http://en.wikipedia.org/wiki/Inductor #1 - It might help to view a perfect coil first. When you put a V on a coil, it sees it as an infinite R and as it 'charges', the I goes to max (infinite current causes problems ;) . The actual shape of the 'ramp up' depends on the application (waveform) V - sine, linear ramp, step, etc. Obviously, the voltage and current are not in sync during this time until saturation. It's often described as a reverse EMF too. Another way to look at it is to think separately at the real resistance of the wire, and the imaginary resistance created by the magnetic field, which goes away as it saturates. For a real coil, think of it as a perfect coil and resistor in series. At t=0, the coil is a very high 'R' and much higher than the real R. As it saturates, the imaginary R of the coil goes to 0 and the real R predominates (at DC, a coil is just a wire with resistance R). The first link has some animations. #2 - Actually, thinking of a capacitor, then taking the 'opposite' is a good way to look at things. With a cap, the E field is a voltage, and the open circuit presented by the 'disconnection' is an infinite resistance, so no current flows. In a coil, the collapsing magnetic field generates a current, which then runs to whatever resistance is across it. You can't stop the collapsing field, so you can't contain it. If it's some finite load R, then E=IR, with I being a function of time. If, you just open a switch or 'pull the plug', you get a spark. That's E getting very big because R is air gap resistance and the voltage becomes greater than the breakdown voltage of the medium and the distance (air in this case) and jumps the gap. To be exact, there will be a very small dissipation in the R of the coil wire too, but it is many orders of magnitude smaller dissipation. Remember, v=L(di/dt) which means it depends on the 'waveform' (even if only 1/2 or 1/4 of a cycle), which means it depends on the time and way V changes. For a switch, it's a step function. Imagine that all for a minute, and do the 'opposite' imagining with a cap, which you seem to know already. Then visit those web sites. See if that process doesn't help put it together for you. ** note: for the purists, the discussion contains the usual 'perfect' components found in academia, and may not be suitable for all audiences. Leakages, et al, are excluded and not needed at this level... ;) -Skip On 3/4/2010 3:19 PM, Electron wrote: > > Dear friends, > I'd like to get a better insight into inductors by asking two questions, > hoping it may be of general interest: > > 1) As we know very well, once we apply power to an inductor the current will > build up till it saturates. At that point, the current we make flow into the > inductor dissipates as heat by the well known joule-effect formula Q=I^2*R*T > Now my first question is: BEFORE we reach coil saturation, does the above > formula still apply, maybe substiting R with the instantaneous impedance, > or part of the energy won't become heat but instead will be stored into > magnetic energy? Apologies if I don't express myself like an engineer: I am > not of course, but I am very passionated about the matter. > > 2) If part of the energy is temporarily "lost" (stored!) into a magnetic field, > imagine if we disconnect completely the inductor in 0 time.. how long will the > energy be retained, and what will estiguish it? > An analogy with a capacitor is almost impossible, as (internal losses set aside) > it will retain the stored energy indefinitely. However an inductor will lose it > pretty quickly, that's my experience at least. > > Thank You, > Mario > -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist