# Confusion of how current can flow in a 1+ henry LR circuit

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?

• electronics.stackexchange.com/questions/288380/… – G36 Jan 1 at 1:42
• 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 – Chu Jan 1 at 1:56
• 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? – csteifel Jan 1 at 1:56
• 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. – Chu Jan 1 at 1:59
• I starts at 0 with dI/dt=V/L – 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.

• Play with this. It's safe tinyurl.com/yden3c7p – Sunnyskyguy EE75 Jan 1 at 2:27
• 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. – glen_geek Jan 1 at 2:45
• 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. – Sunnyskyguy EE75 Jan 1 at 2:55
• Oh, that cries out for a demonstration on YouTube! – TimWescott Jan 1 at 17:13
• 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. – 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}\$$.)

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}}\$$.
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.