eli5: why is the voltage across an inductor different from ohm’s law?

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it states that the voltage across an inductor is V = L(di/dt), why do we not use V = IR? Or what is the derivation of V = L(di/dt) or the conceptual explanation?

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5 Answers

Anonymous 0 Comments

The dI/dt is the rate of change of current over time. That means current has to be changing, which only happens in an AC circuit.

In an AC circuit you have to.worry about reactance, which capacitors and inductors have (and still has units of Ohms). For a resistor, the reactance Xr = R, so Ohm’s law holds. For an inductor, the reactance is Xp = ωL, so it’s not longer Ohmic and you cant use Ohm’s law. (ω is angular frequency, and equals 2πf). It also applies to a capacitor, Xc = 1/Cω

To find the total reactance of a circuit, you would add the reactances of the resistors, inductors, and capacitors independently just like you would in any other circuit so you’re left with 3 values. One for resistors, one for inductors, and one for capacitors

You can then find the reactance of the whole circuit with X = sqrt(Xr^2 + (Xl – Xc)^2 )

Anonymous 0 Comments

Ohms law is about DC system whre you have a steady state. That is the common ohms law with just real variables.

If you have AC system you need to include capacitive and inductive effects. The result of them it that the change of voltage is not in contact with the whole system and that the current is not identical in all of the wires.

If you instead of resistance use impedance Z, that is a way to describe both resitive, capacitive and capacitive effects you can still use ohwm law.

The variable is then a complex number.

For an inductacne the impedance is Z = jwL where j is used instead of I for the imaginary unit. w is the frequency and L is the inductance.

V = I * Z = I* JwL

That is the equation you can use for an inductance. It is relatively simple for just a sine wave when w is a single frequncy.

Anonymous 0 Comments

Ohms law only applies to *Ohmic* devices.

That is devices where I increase proportionally to V across the whole range. Most resistors fall under this and we treat incandescent light bulbs and heating elements like they are because they’re close enough in the operating range, but a cold incandescent lightbulb has about 1/10th the resistance of a hot one so the current changes significantly as voltage increases

For non-Ohmic devices you have to use the right equation. For inductors its V=L di/dt, for capacitors its I=C dv/dt, for diodes and transistors its a lot more complicated

The important thing is knowing what equation to apply when. Just as you can’t solve projectile motion with just Distance=Velocity*Times you need to use all of the *proper* equations to get to the right result in electronics

For an inductor, the current flow induces a magnetic field and magnetizes the core. If you crank current up quickly you try to quickly change the magnetic field in the core and the existing magnetic field pushes back on that change causing the voltage required to get that change in current to go up. If you go slowly then everything has time to get into the right alignment and you can build up to the same magnetic field with much lower voltage, or once you’re fully charged up the push back from the inductor drops to zero and now its basically a short.

V=L di/dt is a fundamental relationship, it is a first principle and therefore cannot be derived from other lower level things

Anonymous 0 Comments

Because Ohm’s law is not universal. It is a special case, that works for some electrical devices, and doesn’t work for others.

V = L(di/dt) is derived from experiment – you can take an inductor, run current through it, measure it, and see that it matches V = L(di/dt).

Ohm’s law is kinda like Aristotle’s law of motion: F=mv. It works, if there is a friction (moreover – a particular kind of friction). But if a friction is 0, or of a wrong kind – it breaks.

There are other devices that do not follow Ohm’s law. In semiconductors, the current is limited not by friction, but by an amount of free electrons – so they can have current saturation – when you increase voltage, but the current stays the same. Superconductors and vacuum have no friction at all – so they can have current without voltage.

Anonymous 0 Comments

There are many good answers here, withthe math all worked out, but I’d like to add an analogy: in capacitors and inductors, the apparent resistance of the component depends on what it is actually doing at the moment, and what it’s doing depends on what happened to it earlier.

An empty capacitor can pass loads of current, because there’s no voltage in it pushing back, so it’ll start out acting as a wire, then act more as a resistor as it’s being charged, until it’s full and acts like an open circuit. When it’s being discharged, it’ll act like a tiny battery that quickly runs out, before returning to wire->resistor->open circuit.

An inductor does the exact opposite: as soon as power is applied, a magnetic field is created which induces an opposing voltage that tries to eliminate the current flowing. If it got eliminated completely the field would stop changing, and only changing fields can cause induction, so there’ll be a balance where the field is slowly changing and the current slowly increasing, just as if there was a resistor with gradually diminishing resistance.

But just like a capacitor, the inductor stores the energy being put in before reaching steady state, so if you try to diminish the current, the magnetic field will fight this as well, now by generating more current as it winds down.

Fun fact: you can create high currents by shorting out a fully charged capacitor, or high voltages by suddenly disconnecting a fully energized inductor. These are both very useful phenomena, as well as fun to play around with.