Average back emf induced in coil

1. Feb 2, 2016

lloyd21

1. The problem statement, all variables and given/known data
The coil in a loudspeaker has an inductance of L = 56uH (or 5.6 x 10^-5 H). To produce a sound frequency of 20 kHz, the current must oscillate between peak values of +2.2 A and -2.2 A in one half of a period. What average back emf is induced in the coil during this variation? How does this compare to the applied emf of 18V?

2. Relevant equations
T = 1/f

emf = L(ΔI/Δt)

L = N(ΔΦB/Δl)

3. The attempt at a solution
back emf = (5.6 x 10^-5)[+2.2 - (-2.2)/Δt]

My problem here is I do not know where I can obtain Δt. I am assuming that it has something to do with the frequency of 20kHz in one half period (1/2T).

T = 1/f --> 1/2T = 1/f --> 2/f -- > T = 2/20000 = 1.0 x 10^-4 secs

If this is the case then, by subbing, Δt = 1.0 x 10^-4 secs, into the above equation, I get:

back emf = (5.6 x 10^-5)[+2.2 - (-2.2)/1.0 x 10^-4 secs]
= (5.6 x 10^-5)[4.4/1.0 x 10^-4 secs]
= (5.6 x 10^-5)(44000)
= 2.464 V

Is this correct?

2. Feb 2, 2016

lloyd21

Okay, so if I am considering a whole period than my currents are actually +4.4 and - 4.4. Additionally, my period is actually just T = 1/f = 1/20000 = 5.0 x 10^-5.

back emf = (5.6 x 10^-5)[+4.4 - (-4.4)/5.0 x 10^-5 secs]
= (5.6 x 10^-5)[8.8/5.0 x 10^-5 secs]
= (5.6 x 10^-5)(176000)
= 9.856 V

3. Feb 2, 2016

Staff: Mentor

So far I'm only seeing work copied from another poster, including their errors. What is your understanding of how the problem should be approached having read through that other[/PLAIN] [Broken] thread?

The other[/PLAIN] [Broken] thread included a discussion of assumptions made about the form of the signal being applied. What assumptions will you make for your attempt?

Is the current increasing or decreasing over the time period in question? What does that imply about the sign of $ΔI/Δt$ and the resulting back-emf?

Last edited by a moderator: May 7, 2017
4. Feb 2, 2016

lloyd21

This thread is the exact question I have, however I approached it differently, and this is what I got:

n = 2(pie)fL = 2(pie)(20x10^3Hz)(5.6x10^-6H) = 7.04 (ohms)

Then Irms = I/sqrt(2) = 2.2 A / sqrt(2) = 1.56 A

Then Vrms = Irms x n = 1.56A x 7.04 (ohms) = 21.9V

5. Feb 2, 2016

Staff: Mentor

It's a valid approach, but you'll need to apply some thought to how to interpret the results in light of the specific time period defined by the problem, and make sure that your result is an average value not an rms value.

Your Vrms value calculation result is double what it should be. Check your calculation.

Note that you don't need to convert to rms values for the calculation. It's fine to use peak values and get a peak result. Then you can translate the peak voltage to an average voltage for the given half-cycle.

6. Feb 2, 2016

lloyd21

So the first approach is much more accurate ?
back emf = (5.6 x 10^-5)[+4.4 - (-4.4)/5.0 x 10^-5 secs]
= (5.6 x 10^-5)[8.8/5.0 x 10^-5 secs]
= (5.6 x 10^-5)(176000)
= 9.856 V

7. Feb 2, 2016

Staff: Mentor

It's not more accurate. It's just another approach. There should be some justification as to why the overall change in current for the half-cycle leads to an average back-emf over the half-cycle: It's the equivalent of assuming a straight-line (ramp) shape for the current waveform, rather than a sinusoid. So a constant dI/dt over the period and thus a constant back-emf. Why should that work?

Regarding your calculations, why have you taken the time period to be a full cycle, and why have you doubled the current values?

As you've written it you have a final current of +4.4A and an initial current of -4.4A. Shouldn't that be an initial current of +2.2A and a final current of -2.2A? And the period of the half-cycle should be 2.5 x 10-5 s, right?

Your reactance approach is interesting because it assumes a sinusoidal signal, and there's a simple relationship between the peak value and the average value for a half cycle of a sine curve. So no special justifications are involved.

8. Feb 2, 2016

lloyd21

sorry the initial attempt was
back emf = (5.6 x 10^-5)[+2.2 - (-2.2)/Δt]

T = 1/f --> 1/2T = 1/f --> 2/f -- > T = 2/20000 = 1.0 x 10^-4 secs

If this is the case then, by subbing, Δt = 1.0 x 10^-4 secs, into the above equation, I get:

back emf = (5.6 x 10^-5)[+2.2 - (-2.2)/1.0 x 10^-4 secs]
= (5.6 x 10^-5)[4.4/1.0 x 10^-4 secs]
= (5.6 x 10^-5)(44000)
= 2.464 V

9. Feb 2, 2016

Staff: Mentor

You've incorporated the calculation error from the original thread again. The period of one cycle at 20kHz is 5 x 10-5 seconds. Half of that is 1/(2f) = 2.5 x 10-5 s.

And you still haven't processed the fact that the current is specified to be decreasing over time for the half-cycle described, so $ΔI$ should be negative.

If you're going to copy another's work, at least check it for accuracy Are you just looking for someone to verify a result you can copy?

10. Feb 2, 2016

lloyd21

I just dont know what to do after that thread ended. If its wrong completely or if it was solved and thats why no one else responded haha. Ive been stuck on that and another question for about a week and a half!

11. Feb 2, 2016

Staff: Mentor

You could work to a solution of the problem yourself here. But you should show your own work, not someone else's work, particularly if you're not understanding it sufficiently to avoid copying their errors. Copying something without understanding it won't help you in the long run.

Make a clean start and pick an approach. I have a feeling that the problem author intended a straight application of $E = L \frac{ΔI}{ΔT}$, where you were to determine the time interval and change in current for that particular interval. A more complex version of this might use calculus to find the actual dI/dt, and average that over the time interval.

But your reactance approach is, in my opinion, very clever and avoids much of the hand-waving you'd have to do to justify the first approach (waveform shape). The only tricky bit involves determining the average value of a half-cycle of a sinewave from its peak value. But that's a very simple bit of calculus, and as a standard result could be simply quoted from a table.