I'm trying to understand how current actually flows in a circuit where you have a 1 Henry or greater inductor in it.

I understand that the voltage across the inductor is V = L*(di/dt) but if L is 1 or greater then how do actually get a starting current flow at all assuming say a 1 ohm resistor with a DC power source of 10 volts, making i(max) = 10 amps.

Wouldn't as soon as you connect up an "ideal" dc power source you get an instantaneous change of 10amps over lets just say 1 time unit (however small you want to choose that to be) and therefore back emf be 10v and then the current would be 0 since those two voltages would cancel each other out? And therefore the change in current couldn't actually be possible unless di/dt were less than i(max), but even then if you have a greater Henry inductor even that doesn't hold true? Does L*(di/dt) always HAVE to be less than power supply voltage, is this situation just not possible, if so why?

  • 1
    \$\begingroup\$ electronics.stackexchange.com/questions/288380/… \$\endgroup\$ – G36 Jan 1 at 1:42
  • 1
    \$\begingroup\$ A couple of things, here. The instantaneous current at t=0 is not 10 A - the rate of change of current is 10 A/s. Also, there must be a current, because it's increasing! If it weren't there would be no di/dt \$\endgroup\$ – Chu Jan 1 at 1:56
  • \$\begingroup\$ But that would also mean that even in a 100% ideal world di/dt could never be equal to i(max) which in an ideal world shouldn't it be since the inductor shouldn't have a resistance? \$\endgroup\$ – csteifel Jan 1 at 1:56
  • 1
    \$\begingroup\$ How can di/dt be the same as i? They're different things. di/dt = 0 when 10 A flows. A current can have a rate of change when it's zero, which is exactly what's happening here. \$\endgroup\$ – Chu Jan 1 at 1:59
  • 1
    \$\begingroup\$ I starts at 0 with dI/dt=V/L \$\endgroup\$ – Sunnyskyguy EE75 Jan 1 at 2:28

No there is no instant current.

I starts at 0 with dI/dt=V/L when you connect a battery across a large coil regardless of Rs.

I put an 18650 Li Ion battery directly on a 22 Henry inductor ( the primary of a 5MVA transformer) intended for 30kV. The result was measured with a DMM DC current mode.

It was, as expected, with 3.8V across it starting from 0A and rising slowly at dI/dt=V/L= 138 mA per second. After a minute , I disconnected and got a nice long arc around 8 Amps that lasted for a while. The ESR of the arc depends on the current and the current density at contact, but I had stored a lot energy.

  • \$\begingroup\$ Play with this. It's safe tinyurl.com/yden3c7p \$\endgroup\$ – Sunnyskyguy EE75 Jan 1 at 2:27
  • \$\begingroup\$ Shorting big capacitors, and opening big inductors sounds like "component abuse", Tony ;-) When I did that demo, it was with a weak little 1.5V battery, and I snubbed its ends between my fingers - still made me jump. \$\endgroup\$ – glen_geek Jan 1 at 2:45
  • \$\begingroup\$ Well this battery could handle 10A but it started to get warm. and the transformer could handle 5MW @ 30kV, so I barely tickled it. yet I didn't let it zap me. \$\endgroup\$ – Sunnyskyguy EE75 Jan 1 at 2:55
  • 1
    \$\begingroup\$ Oh, that cries out for a demonstration on YouTube! \$\endgroup\$ – TimWescott Jan 1 at 17:13
  • \$\begingroup\$ I should have thought of that at the time. You can make a nice steady corona with the right holding current with a tungsten rod and right gap with just a Li Ion cell. But the transformer had high L/R ratio and weighed many tons and was much taller than I. But I was solving an H2 gassing crisis slowly dissolving in the oil tank. \$\endgroup\$ – Sunnyskyguy EE75 Jan 1 at 17:26

I believe that your difficulty springs from the fact that you are trying to compare current with the rate of change of current. You cannot do this -- they are different things. It's an apples = oranges sort of fallacy.

\$\frac{di}{dt}\$ is how fast current is changing so it's related to current over time, but at any one instant, \$i\$ and \$\frac{di}{dt}\$ are, by themselves, independent.


Central question

I think your question may center on this:

and therefore back emf be 10v and then the current would be 0 since those two voltages would cancel each other out

I think you are curious why, if the back-EMF is the same as the applied EMF, things don't just "cancel out" and therefore why does any change take place, at all.

Short overview

An inductance is defined by its design and implementation. Just as a capacitance is defined by its design and implementation. Capacitors hold charge, \$q\$, and are defined such that \$C=\frac{\text{d}\,q}{\text{d}\,V}\$ (or, the infinitesimal change of charge with respect to some infinitesimal change in voltage.) Inductors hold flux, \$\phi\$, and are defined such that \$L=\frac{\text{d}\,\phi}{\text{d}\,I}\$ (or, the infinitesimal change of flux with respect to some infinitesimal change in current.)

You can think of the flux (\$\phi=L\cdot I=\int V\:\text{d}\,t\$) of an inductor, in Webers, as being the dual of charge (\$q=C\cdot V=\int I\:\text{d}\, t\$) on a capacitor, in Coulombs. In mechanical physics, these are the equivalent of momentum: \$p=m\cdot v=\int F\:\text{d}\,t\$.

The equivalent of an external mechanical force (\$F\$) for an inductor is voltage (\$V=\frac{\text{d}\,\phi}{\text{d}\,t}=L\,\frac{\text{d}\,I}{\text{d}\,t}\$) and for a capacitor it is current (\$I=\frac{\text{d}\,q}{\text{d}\,t}=C\,\frac{\text{d}\,V}{\text{d}\,t}\$). The equivalent for mechanical velocity, \$v\$, for an inductor is current, \$I\$, and for a capacitor it is voltage, \$V\$. (The equivalent for mechanical acceleration, \$a\$, for an inductor is \$\frac{\text{d}\,I}{\text{d}\,t}\$ and for a capacitor it is \$\frac{\text{d}\,V}{\text{d}\,t}\$.)

An answer

I'd earlier written that \$L=\frac{\text{d}\,\phi}{\text{d}\,I}\$. But this is also \$L=\frac{\text{d}\,\phi}{\text{d}\,I}=\frac{\frac{\text{d}\,\phi}{\text{d}\,t}}{\frac{\text{d}\,I}{\text{d}\,t}}\$.

The inductance is then also defined as the rate of change of flux with respect to the rate of change in its current. A changing current in the inductor implies there must be a changing magnetic flux. A changing flux yields a non-Coulomb electric field that curls around the region of changing flux in the inductor. This non-Coulomb electric field sets itself up to oppose the changing current and is called back EMF. This back EMF must be met by exactly the same magnitude of applied EMF (the forces are equal, but opposite, just as in mechanical physics every force has an equal but opposite force) in order to yield some rate of change in the flux (and current.) The inductance value itself determines the ratio.

In some sense, you can think of the back EMF as similar to the inertia of mechanical mass. An inductor is like mass and when you apply a force (here, a voltage \$V\$) to it, it responds with an equal but opposite force (here, a back EMF \$\epsilon\$.) It's just the response to applying a force on the inductor and it must be equal, but opposite. So there is no conflict where these voltages "cancel out and nothing happens." Instead, back EMF, \$\epsilon\$, is simply the inertial counter-force that opposes some applied external force, \$V\$, on the inductor.


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.