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Inductor opposing change in current. 
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#1
May713, 08:33 PM

P: 746

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
May713, 09:48 PM

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#3
May713, 09:50 PM

P: 746

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


#4
May713, 10:09 PM

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P: 41,317

Inductor opposing change in current.
[tex]V(t) = L \frac{dI(t)}{dt}[/tex] 


#5
May713, 10:12 PM

P: 746

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
May713, 10:50 PM

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P: 3,954

An analogy which is really quite good if you compare the equation with Newton's [tex]f=ma=m\frac{dv}{dt}[/tex] 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
May813, 12:49 AM

P: 32

Woopydalan,
Your equation is correct. It holds true for selfinductance. 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, selfinductance (L) becomes zero. Thus, current flows. This is the simplest and only explanation. 


#8
May813, 01:02 AM

P: 1,506

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
May813, 05:37 AM

P: 533

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. 


#10
May1013, 09:12 PM

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#11
May1013, 09:19 PM

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P: 12,074

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
May1113, 08:19 PM

P: 746

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
May1213, 01:43 AM

P: 32

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
May1213, 03:54 AM

P: 533

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: [tex] V = L \frac{\mathrm{d}i}{\mathrm{d}t} \Leftrightarrow \frac{\mathrm{d}i}{\mathrm{d}t} = \frac{V}{L} [/tex] 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
May1213, 06:27 AM

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P: 12,074

I'll give a short and simple answer.



#16
May1213, 08:02 AM

Sci Advisor
Thanks
P: 2,546

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


#17
May1213, 09:35 PM

P: 746

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
May1313, 10:56 AM

P: 1,506

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. 


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