Inductor considered as a broken wire.

[Sefrez]

I quote from my book:
"Just after the switch is closed, the inductor acts to oppose a change in the current through it."

"Because the current through each inductor is zero before the switch is closed, it will also be zero just afterward. Thus, immediately after the switch is closed, the inductors act as broken wires, as indicated in Fig..."

What does that supposed to mean, "Because the current through each inductor is zero before the switch is closed, it will also be zero just afterward."?

I can say contradiction for one. If there is "no" current right afterward, the is no opposing of current.

So I am not sure if I am missing something but that sounds like a poor analogy to explain what is going on.

If they are said to have no current through them, then you might as well agree that any other point in the circuit is of the same, zero current - "right" after the circuit is closed.

So going by that description, I don't understand how you can treat inductors as broken wires at first.

So my question is, why can they be considered as broken, considering my book has a terrible explanation?

Does it have something to do with the resisting induced emf? Though going by the book, it "has no current through it" so it might as well not be an inductor as an inductor is ONLY an inductor if there is a change in current through it, otherwise it is just a wire!

An help is appreciated.

 

[gneill]

Try a mechanical analogy. Suppose you have a mass M sitting at rest and suddenly apply a force to it. The mass does not immediately jump to some high velocity! Immediately after the force is applied its velocity is still zero but starting to change (accelerating). The associated equation is $F = M\frac{dV}{dt}$.

The inductor has the electrical equivalent of mechanical inertia when it comes to current. The associated equation is $V = L \frac{dI}{dt}$.

Note that components such as (ideal) resistors do not exhibit this phenomenon. The equations that describe them are not time dependent: V = I*R.

 

[Sefrez]

Thanks for the reply.

Try a mechanical analogy. Suppose you have a mass M sitting at rest and suddenly apply a force to it. The mass does not immediately jump to some high velocity! Immediately after the force is applied its velocity is still zero but starting to change (accelerating). The associated equation is $F = M\frac{dV}{dt}$.

Yes, I understand this. My problem is this: my inability to differentiate the inductor from any other wire segment in the circuit when there is zero current though it. Because, in the context of the book quotes, saying that the current in the inductor is zero right after the circuit is closed is like saying it is zero every where else in the circuit. Just as the mechanical analogy describes as you stated. The current simply does not instantaneously reach some surface value.

But with no current through the wires right after the circuit is closed, is not a coil of wire the same as a straight wire (electrically)? They seem the same in one to me. That is where I am confused, and leads me to say, "every segment of the wire can be considered broken at first." And with this, the circuit practically becomes a mess of broken up segments and your back to the idea that the circuit might as well not be called a closed circuit.

The inductor has the electrical equivalent of mechanical inertia when it comes to current. The associated equation is $V = L \frac{dI}{dt}$.

Note that components such as (ideal) resistors do not exhibit this phenomenon. The equations that describe them are not time dependent: V = I*R.

This seems to be getting closer to solving my confusion. So, if the inductor is to be considered broken, does this mean that V = L dI/dt is such that V equally opposes the potential before the inductor (say by an emf)? That would at least mean that no current flows.

Thanks.

 

[gneill]

Yes, I understand this. My problem is this: my inability to differentiate the inductor from any other wire segment in the circuit when there is zero current though it. Because, in the context of the book quotes, saying that the current in the inductor is zero right after the circuit is closed is like saying it is zero every where else in the circuit. Just as the mechanical analogy describes as you stated. The current simply does not instantaneously reach some surface value.

But with no current through the wires right after the circuit is closed, is not a coil of wire the same as a straight wire (electrically)? They seem the same in one to me. That is where I am confused, and leads me to say, "every segment of the wire can be considered broken at first." And with this, the circuit practically becomes a mess of broken up segments and your back to the idea that the circuit might as well not be called a closed circuit.

Actually, there is a kernel of truth there. Any real length of wire, even 1cm long and perfectly straight, has some inductance. It may be very, very small, but it is still there as evidenced by the fact that all current-carrying wires produce a magnetic field.

So any real circuit will exhibit inductive effects to one extent or another. This will include the "zero current at the instant the switch is closed" effect. It's just too small to notice (unless you look really closely!) when the inductance is vanishingly small, or when stray capacitance between wires, components, and the environment make the resulting impedance more capacitive than inductive.

These tiny stray inductances and capacitances give rise to what are called "transient phenomena", which show up as short-lived transitions or oscillations (ringing) in circuits when they are "hit" with sudden potential changes. You'll learn all about these "second order" effects when you study circuits with both inductance and capacitance (LC or LRC circuits).

When we draw a circuit using ideal wires and components the wires are assumed to have no resistance, inductance or capacitance, and all such characteristics are lumped into individual ideal components.

This seems to be getting closer to solving my confusion. So, if the inductor is to be considered broken, does this mean that V = L dI/dt is such that V equally opposes the potential before the inductor (say by an emf)? That would at least mean that no current flows.

Yes, in effect the inductor produces an emf in reaction to the rate of change of current passing though it. An applied (external) voltage V_x causes a dI/dt = V_x/L, and the resulting dI/dt in the inductor creates a changing magnetic field in its loops. But a changing magnetic field induces an emf in a loop of wire! So a potential difference appears across the inductor as a result of any current change. As it happens, this potential always arises with a polarity that opposes the change in current. So it's the electrical equivalent of the inertial force that arises when you apply a force to a mass.