# Inductor opposing change in current.

1. May 7, 2013

### Woopydalan

Hello,

I'm wondering, if you have an inductor and resistor that has the switch open, so current is zero. Then you turn on the switch, the inductor is going to resist this change, it wants to be zero, right? Therefore, why doesn't the current indefinitely stay as zero, since the inductor would like to be at that value that it was at initially.

2. May 7, 2013

### Staff: Mentor

Can you show us the equation that relates current and voltage for an inductor?

3. May 7, 2013

### Woopydalan

ε = -L di/dt

So if it opposes a change (di/dt = 0) then ε is 0, then no current

4. May 7, 2013

### Staff: Mentor

There is no negative sign, first of all. And the equation should show the current as a function of time for an applied voltage:

$$V(t) = L \frac{dI(t)}{dt}$$

5. May 7, 2013

### Woopydalan

Yeah, so my conceptual understanding of an inductor is that it would like to maintain the same current through it. If that is the case, why does it ever change? I guess unless something can overcome the resistance that the inductor puts forth to resist that change. What is it?

6. May 7, 2013

### Staff: Mentor

Look at the equation berkeman posted again. L is non-zero, V is non-zero, therefore dI/dt must be non-zero. And if dI/dt is non-zero at some time t, then even if I is zero at time t it won't be zero a moment later; the voltage is pushing current through the inductor.

An analogy which is really quite good if you compare the equation with Newton's $$f=ma=m\frac{dv}{dt}$$

Because of inertia, a mass hanging from a rope resists changes in speed unless you apply a force to it. If it's not moving and you push on it, it resists. But it will start swinging, and once it does it will resist any attempt to stop it.

7. May 8, 2013

### wirefree

Woopydalan,

Your equation is correct. It holds true for self-inductance.

You are right about the inductor resisting current. It holds true because of Lenz's Law.

As for your actual question on why the current doesn't remain zero: it is because, in steady state, flux through the inductor is zero. Hence, as stated by Faraday's Law, induced emf which opposed the current in the first instance eventually disappears, that is, self-inductance (L) becomes zero.

Thus, current flows. This is the simplest and only explanation.

8. May 8, 2013

### technician

My text books show induced emf = -d∅/dt. = -L.di/dt
The minus sign is taken to be an indication of Lenz's law.
Where is my text book (or my reading of the book!) going wrong

9. May 8, 2013

### milesyoung

It resists changes in current.

If the current through the inductor is the only source of flux linking its turns, then the flux through the inductor is only zero when the current through it is zero.

If by steady state you mean the rate of change of flux will tend to, or become, zero as t->∞, then you're assuming something which isn't true in general, e.g. it's not true for a constant voltage applied across an ideal inductor.

If the inductor induces an emf, then the rate of change of current through it must be nonzero, i.e. the magnitude of the current changes with time. You have assumed that which makes your conclusion true.

10. May 10, 2013

### Redbelly98

Staff Emeritus
The negative sign is usually included in intro physics textbooks, and omitted in electrical engineering books. But the two use a different convention for the direction of the current in an inductor, relative to the potential difference. They are equivalent descriptions.

11. May 10, 2013

### Redbelly98

Staff Emeritus
If that were true, we would not describe an inductor with that equation. We would simply say ε=0, i=0, and be done with it.

You can think about it this way:

If there is no current in the inductor, then there's no current in the resistor either.
In which case, there is no voltage difference (V=iR) across the resistor.
So then, the inductor must get the full battery voltage ε.
Therefore, di/dt = ε/L is not zero.

12. May 11, 2013

### Woopydalan

I guess my question is that if an inductor resists changes to current, if the inductor had 0 current in it, why would it have current in it if the switch is closed? Wouldn't it resist having a nonzero value?

13. May 12, 2013

### wirefree

When the switch is closed, the current and the resulting magnetic field increases from zero to the maximum value in a short time. This results in a increase in magnetic flux through the inductor and, hence, again induces an electric current in the inductor.

The above forms the basis of Faraday's Law which was observed during experiments conducted around 1830.

14. May 12, 2013

### milesyoung

If the inductor didn't oppose changes in current, you would be able to make the current through it "jump" instantaneously from one value to another, like you could for an ideal resistor. At the other end of the spectrum, it would oppose any and all changes to current and the rate of change of current through it would always be zero.

If you apply a voltage V across an ideal inductor with inductance L, you have from Faraday's law:

$$V = L \frac{\mathrm{d}i}{\mathrm{d}t} \Leftrightarrow \frac{\mathrm{d}i}{\mathrm{d}t} = \frac{V}{L}$$
Clearly then, the rate of change of current through the inductor can be nonzero. If you imagine plotting the current through the inductor as a function of time, you'll notice that V/L gives you the slope of this line. Decreasing the inductance L would give you a larger and larger slope, i.e. the change of current through the inductor would occur more rapidly.

You might say the inductance is a measure of "how much" the inductor resists changes in current.

15. May 12, 2013

### Redbelly98

Staff Emeritus
I'll give a short and simple answer.

Resisting change means that the current does not (nearly) instantaneously jump to a nonzero value, as it can in a resistor. Think about what happens in your circuit if there is no inductor.

16. May 12, 2013

### vanhees71

Perhaps it's easiest to consider the most simple problem of this kind analytically. Take the usual assumptions of quasistationary circuit theory, i.e., the extensions of the whole circuit is much smaller than the wavelength of typical em. waves of the problem, so that the displacement current and retardation can be neglected. Then everything is described by linear ordinary differential equations and the circuit is described by resistance, capacitance and self-inductance.

We consider a coil of self-inductance L in series with a resistor, attached to a DC battery. We switch on the circuit at $t=0$. From Faraday's Law we find
$$L \dot{i}+R i=U(t),$$
where
$$U(t)=U_0 \Theta(t).$$
First we solve the homogeneous equation
$$\dot{i}=-\frac{R}{L} i.$$
Obviously the general solution is
$$i(t)=A \exp \left (-\frac{R}{L} t \right).$$
Then we need one special solution of the inhomogeneous equation. Obviously this is given by the ansatz $i_0=B =\text{const}$. Plugging this in the equation, we get for $t>0$ that $B=U_0/R.$ The general solution of the full equation thus is for $t>0$
$$i(t)= A \exp \left (-\frac{R}{L} t \right) + \frac{U_0}{R}$$
Since $i(0)=0$ we finally can determine $A$ and then get
$$i(t)=\frac{U_0}{R} \left [1-\exp \left (-\frac{R}{L} t \right ) \right ].$$
As you see the current won't flow immediately at full strength but only approaches exponentially the static value $i_{\infty}=U_0/R$.

17. May 12, 2013

### Woopydalan

I would like to thank you all for contributing to this thread. I think I am now fully convinced that an inductor can resist change, which doesn't mean that it completely prevents the flow of changing current, much like a resistor doesn't prevent the flow of current entirely.

18. May 13, 2013

### technician

There is something very strange here !!! usually introduced in intro physics textbooks !!!!
Why do some introduce it and others do not, which ones do and which ones don't. The hyperphysics site, often quoted here....what do they do??
I did not know that electrical engineers had a different convention for electric current...how do they cope with kirchoffs laws when communicating with physicists?
An emf is generated across an inductor....not a pd....is there some confusion here about which way current flows.
Your statement does not help anyone wanting to learn physics....especially those using text books.
The minus sign is a recognition of Lenz's law. It is as important in electrical studies as the minus sign is in the analysis of SHM in mechanics. I doubt if anyone would claim that a minus sign is not necessary in SHM and that some intro physics text books decide not to use it.
It is PHYSICS.

19. May 13, 2013

### Redbelly98

Staff Emeritus
Maybe getting off topic here, but I am compelled to defend myself against some criticism of what I posted earlier.
SHM is different. You define displacement to be positive in some direction, and it's understood that's the positive direction for the first and second derivatives and restoring force as well. But when defining positive current to go with a positive voltage difference, it can -- and in practice does -- go either way.

People can -- and do -- define positive current flow to be in either direction: either entering the positive terminal of a device, or entering the negative terminal. For most devices (resistors, batteries, diodes), there is pretty universal agreement. Not so for inductors:

If you're not familiar with the terms "active sign convention" and "passive sign convention", try a google search and learn about them. To be very brief

Active sign convention, V = -L di/dt

Passive sign convention, V = +L di/dt

20. May 13, 2013

### Staff: Mentor

Thanks RB, that's interesting.

21. May 14, 2013

### vanhees71

There is no problem with the direction of currents when going back to the local form. There you deal with vectors, and they have a well-defined meaning of direction. The current density $\vec{j}$ describes the flux of electric charge per unit time per unit area. The total current (charge per unit time) through an area element $\mathrm{d} \vec{F}$, where this surface-normal vector defines the orientation of the area is given by
$$\mathrm{d} \vec{F} \cdot \vec{j}.$$

When dealing with Faraday's Law, you have
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.$$
For a circuit where all wires, resistors, capacitors, and inductors are at rest, this translates into
$$\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{\mathrm{d} \Phi}{\mathrm{d} t}.$$
Here the magnetic flux is defined by
$$\Phi=\int_{F} \mathrm{d} \vec{F} \cdot \vec{B}.$$
The relative orientation of the surface and the boundary is (by convention) according to the right-hand rule.

Keeping this origins of Kirchhoff's Laws in mind, there can never occur problems with the signs of EMF's. One has just to keep track of the relative-sign conventions for the surface and its boundary according to the right-hand rule.

22. May 14, 2013

### technician

I see what you are saying and the conclusion is that e = -Ldi/dt and e = +Ldi/dt are both acceptable ??
Presumably a sign convention must be stated every time these expressions are used? Otherwise there will be total confusion.
The reference to SHM was to highlight that -signs do have a significance elsewhere.
It is not beyond comprehension that someone would be able to come up with a sign convention for force that made it acceptable to use F=+kx as a definition of SHM. I hope nobody goes to the trouble !!!
Google searches always produce something 'interesting'. I prefer using text books and doing text book searches.

23. May 14, 2013

### vanhees71

I don't understand these drawings cited in the previous posting. This must be some convention used in electrical engineering which might help to keep track of the correct relative signs.

It's also a bit hard to explain without drawings. As an example take a ideal inductor in series with a resistor connected to an AC voltage $V(t)$. Label the AC voltage with arbitrary + and - signs at your choice. We assume that the voltage measured in this sense is given by $V(t)$. Then take the current in a direction pointing away from the pole labeled +. Now you integrate Faraday's Law around the circuit in the sense of the current. The line integral gives
$$\int_{C} \mathrm{d} \vec{r} \cdot \vec{E}=R i - V(t).$$
Now according to the right-hand rule for the induction of the magnetic field in the wire in connection with Ampere's Law together with the right-hand rule for the orientation of the surface integrated over, enclosed by the circuit gives you $\Phi=+L i$. Thus according to Faraday's Law you get
$$R i-V(t)=-L \frac{\mathrm{d}i}{\mathrm{d} t}.$$
No sign problems whatsoever!

24. May 14, 2013

### technician

The last post looks sensible. In the final equation I would write
Ri - E(t) = -Ldi/dt because E(t) is an emf.
Then E(t) -Ldi/dt = Ri. This is a very familiar equation in physics and is the statement that induced emf in an inductor is -Ldi/dt
There are no problems with signs.
Every standard physics text book gives induced emf = -Ldi/dt and one of the aims in these forums is that explanations should conform with standard text books.
The - sign is how Lenz's law is expressed....a change is resisted not enhanced.

25. May 14, 2013

### milesyoung

In circuit analysis you deal with element laws and sign conventions. When you assign reference directions and polarities for the currents and voltages, respectively, you have to follow up with the correct element laws. Consider this circuit:

Vin = VR + VL

For the resistor and inductor, the direction of current and polarity of voltages follows the passive sign convention, so you have to apply the element laws VR = R*i and VL = L*di/dt. The element laws for the active sign convention are VR = -R*i and VL = -L*di/dt.

You focus on sign conventions and element laws so you don't have to worry about applying Faraday's law etc.

Last edited: May 14, 2013