Already answered, just need quick piece of advice

  • Thread starter lloyd21
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Homework Statement



Hi,
I would appreciate any help with the following:

The coil in a loudspeaker has an inductance of L = 112 microH. To produce a sound of frequency 40kHz, the current must oscillate between peak values of + and - 4.4 A in half a period. What average back emf is induced in the coil during this variation?

My main confusion is, why do we need to know the half period thing...I have no idea how to incorporate it into the solution.

Here's what I have now:

Xl = inductive reactance = 2 pi f L = 2 pi 40 x 10^3 x 112 x 10^-6 = 28.15 Ohm.

Then I rms = I/sqrt 2 = 4.4/sqrt 2 = 3.11 A. Am I correct in looking at rms values here?

Then V rms = I rms x Xl = 3.11 x 28.15 = 87.5 V.

Is that all? Is this V rms the average induced back emf they are asking for? Or am I wrong? I'm really confused about this one.

Any help would be much appreciate! Thanks!
Hi equinox2012, Welcome to Physics Forums.

The back-emf created by an inductor depends upon the rate of change of the current flowing through it. Thus for an inductance L,
V=LdIdt

That's why the information about the change in current over a particular period of time was given

I dont know how to change my calculations to the new formula that was given


The Attempt at a Solution

 

Answers and Replies

  • #2
gneill
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It looks like they want you to find the average value of the back-emf over the period of time given. I'd forget the rms values and write the current as an appropriate function of time. Then use the relationship between voltage and current for an inductor to find the back-emf as a function of time, too. Go from there to find the average value. Note: some calculus may be involved...
 
  • #3
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It looks like they want you to find the average value of the back-emf over the period of time given. I'd forget the rms values and write the current as an appropriate function of time. Then use the relationship between voltage and current for an inductor to find the back-emf as a function of time, too. Go from there to find the average value. Note: some calculus may be involved...

Thats where im confused, there isnt suppose to be any calculus. I just dont understand the V=L dI/dt ....it gives me a massive voltage which is not correct
 
  • #4
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I tried gneill and the results defenitley did not come out the way they are supposed to. the second part to the question says - How does this compare to the applied emf of 18V?.......thats why i think the first answer was more accurate? im lost...
 
  • #5
gneill
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No calculus, eh? Then I'll need to rethink what they're after.

Terms like peak, rms, and average have very specific meanings in electronics. When I read the question my take was that they wanted the average back-emf for the part of the current cycle where it goes from a peak of +4 amps to a minimum of -4 amps.

If the current is represented as a cosine function, the back emf will have the form of a sine function (thanks to that dI/dt). Now, the average value of any sinusoid over time is always zero (positive and negative excursions all cancel out). But it can be non-zero over some fraction of a cycle. That's why I latched onto the half-period that they described. For that half period the back-emf would be a half cycle of a sine, which definitely would have a non-zero average value.

Now you say there's a second part where they give the applied emf as 18 V. Is that 18 V peak or rms?

I think if you find the peak value of the back-emf for 40 kHz you'll be doing okay (your same calculations but skip the rms conversion). You can always find the rms value later.

If there's a discrepancy between applied emf and back-emf on the coil, can you think of any reason for it? Hint: what properties do real-life coils, made of real-life materials, have that ideal ones don't? keywords: resonance effects.
 

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