# voltage drop across inductor

Discussion in 'General Electronics Chat' started by gator27, Nov 12, 2008.

1. ### gator27 Thread Starter New Member

Nov 12, 2008
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Under steady state AC voltage and current is there a voltage drop across an inductor? I konw the voltage drop across an inductor is related to the rate of change in current across the inductor but since the current is AC then there is constantly a rate of change?

If I have a series circuit and the inductor makes up 50% of the circuit impedance does this mean that under steady state AC conditions the voltage drop across the inductor will be half the drop across the entire circuit?

2. ### steveb Senior Member

Jul 3, 2008
2,433
469
Yes, if you are talking about steady state sinusoidal signals. If you are talking about DC steady state, then the voltage is zero across an ideal inductor.

For the most part there is a rate of change. The time rate of change of i=I*sin(w*t) is di/dt=I*w*cos(w*t). So the rate of change is also a sinusoid. There are moments when the rate of change is zero.

That's a trick question. If you consider the peak or rms voltage then it is 50%; however if you look at the instantaneous voltage values, then it is not 50 %, unless the components are identical. This is because there are different phase shifts in the voltages relative to the current. In a series circuit, the currents are equal, hence there is different phase shift for the voltages on each element.

3. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,047
295
The reactance of an inductor is= 2*pi*fL, where f is the frequency, and L is the inductance (x). The result is a reactance in ohms, which can be substituted into ohms law by E=IX. The voltage across the inductor will appear just as it would with a resistance equal to x. However the current will lag the voltage by 90 degrees, according to ELI the ICE man.

eric

4. ### mik3 Senior Member

Feb 4, 2008
4,846
63
The voltage across an inductor is

v=Ldi/dt

If your input current is sinusoidal for example then the voltage across the inductor will be a cosine wave and vice versa.

For the last question yes it will be the half if you are considering RMS values.

5. ### gator27 Thread Starter New Member

Nov 12, 2008
6
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I'm not following why the voltage drop would be different when looking at rms vs instantaneous? Can you explain.

Lets say I have a 480V source in series with a X=50ohm inductor and X=50ohm resistor. Since the total Z value is 100ohm and the inductor is therefore 50% of the total impedance would I have half the voltage of 240V dropped across the inductor?

6. ### steveb Senior Member

Jul 3, 2008
2,433
469
Yes, you are correct that the RMS voltage is 50%, and that's the usual answer. However, if you look at the time dependent voltages across both components, you would find that there is a phase shift between them. The inductor has a 90 degree phase difference between voltage and current, while the resistor has zero degree phase difference. So, there will be instants in time where the resistor voltage (and current) is zero, since this is the nature of an AC signal. At such times, the inductor voltage will not be zero due to the phase difference.

7. ### gator27 Thread Starter New Member

Nov 12, 2008
6
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OK I get it now. So if I'm using an inductor as a current choke device during fauls it will sill have a voltage drop across it (RMS) during steady state conditions on the circuit.

Also If I'm told that a reactor/inductor has to be 50% impedance of the entire circuit then I'm assuming I need to find the total impedance required for the circuit and then size the reactor to half of that value?

8. ### Von Active Member

Oct 29, 2008
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Excellent answer steveb. Shouldn't this be apparent if a two channel scope is used as opposed to two DMM's?

Last edited: Nov 13, 2008
9. ### gator27 Thread Starter New Member

Nov 12, 2008
6
0

So if you read the voltage across the inductor will you read the actual voltage drop across the inductor (in phase) or will you read the phase shifted voltage across the inductor therfore appearing to be more than the actual voltage is.

In other words if I have an 100V source in series with an inductor will I read 100V across the inductor or will I read more than 100V because of the 90deg phase across the inductor?

10. ### steveb Senior Member

Jul 3, 2008
2,433
469
I assume you mean a voltage supply driving an inductor directly with no other components.

If you are measuring RMS voltage with a meter, then 100V will be read if the supply is 100 V RMS. If you use a scope to measure, you will see the entire time depended sinusoidal voltage. The inductor voltage would be in phase with the source because they are the same thing. The inductor current would be phase shifted however.

I'm hoping that I haven't confused you more than helped you. The main point of explaining the difference is that with AC voltages you have to be careful to specify whether you mean RMS, peak, peak-to-peak or instantaneous voltage. Your original question asked if the voltage was 50 %, but you did not say RMS voltage. Even your question above does not specify, so it can be hard to answer the question simply.

11. ### gator27 Thread Starter New Member

Nov 12, 2008
6
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Steve

Thanks for the explanation. I understand the differences you are referring to above. My origonal question was relating to the RMS voltage but after your response I got more interested in some of the other instantaneous readings.

I see what you are saying about reading across the inductor instantaneously. I was thinking that maybe because the current was shifted that this shifted current would cause a shifted voltage drop through the inductor. Therefore when measuring the voltage at the secondar side of the inductor the voltage would be shifted 90deg because it was a result of the impedacne of the inductor times the 90deg shifted current.

I thought then maybe when measuring across the inductor we would be measuring between the primary voltage on the inductor and a 90deg shifted voltage on secondary of the inductor. This I thought would maybe appear to be more than 100V because you would be measuring two out of phase voltages?

12. ### gator27 Thread Starter New Member

Nov 12, 2008
6
0
Thanks for all the help, I now see how the voltage drops and reacts across an inductor.

The one last piece I was trying to figure out, is how an inductor can be used as a voltage choke. I understand how it is used as a current choke but I have been reading that inductors/reactors are used to help eliminate voltage spikes or surges.

Are the inductors actually limiting the voltage spikes themselves or the current associated with them. From what I have read it seems like the inductors are actually reducing the voltage spikes themselves. Can anyone explain in theory why this is?