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How do transformers obey ohm's law?

  1. Jan 16, 2014 #1
    If you apply 100 volts to a 10:1 transformer you will end up with 10 volts out. If you put a 10 ohm resistor on the secondary you will have one ampere passing through the resistor. That's it !!!' Why do I have to apply energy conversion equation to find the intensity it seems meaningless, why don't I just calculate the induced emf then find the intensity by : Induced Emf / R of secondary coil. Please guys, I'm sure that I'm wrong, so I need a good explanation for this
     
  2. jcsd
  3. Jan 16, 2014 #2

    Andrew Mason

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    Because the secondary is a coil, there is inductive reactance in addition to the load resistance. Ohm's law applies to resistance loads. If there is a phase angle between current and voltage, which there will be unless the there is a capacitor in there somewhere, the power is not given by P = VI. It is P = VIcosΘ where Θ is the phase angle between current and voltage. If Θ = 90° (no resistance, only inductive reactance) there is no power consumed at all.

    AM
     
  4. Apr 29, 2014 #3

    Can you detail that a little bit ??
     
  5. Apr 29, 2014 #4

    Philip Wood

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    To a good approximation, the secondary current is given by:

    [tex]I_s = \frac {\epsilon_s}{R_L + R_S}[/tex]
    [itex]\epsilon_s[/itex] = emf in secondary, [itex]R_L[/itex] = load resistance, [itex]R_S[/itex] = resistance of secondary coil.

    The reactance due to self inductance of the secondary coil is, I believe, usually negligible.

    The word you need is 'current', by the way, not 'intensity'.
     
  6. Apr 29, 2014 #5
    I don't understand the difference between both resistances???
     
  7. Apr 29, 2014 #6

    Philip Wood

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    That's easy: the secondary coil of the transformer itself has resistance. So does the 'load' you connect across the secondary coil. These two resistances behave as if they are in series. That's because of the way the emf is induced in the secondary coil.
     
  8. Apr 29, 2014 #7
    How do transformers obey ohm's law??

    So in a step down transformer of turns ratio 2:1 if the voltage in the primary coil is 2V the voltage in the secondary will be 1V

    Then to find the current in the secondary coil we just divide 1volt by the sum of both resistances ????
     
  9. Apr 29, 2014 #8

    Philip Wood

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    Yes. In many cases, though the load resistance is much greater than the coil resistance, so you can forget the coil resistance.
     
  10. Apr 29, 2014 #9
    but I'm confused because this [itex]\epsilon[/itex]p / [itex]\epsilon[/itex]s = [itex]I[/itex]s / [itex]I[/itex]p

    so in my previous example if the current in the primary coil is 1A according to this equation the current in the secondary coil will be 2A not 1A

    that contradicts with what we said before :confused:
     
  11. Apr 29, 2014 #10

    berkeman

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    Please see the "Ideal Transformer" section of the wikipedia entry to see if it clears up your confusion...

    http://en.wikipedia.org/wiki/Transformer

    :smile:
     
  12. Apr 29, 2014 #11

    Philip Wood

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    Let's do an example. Suppose an alternating p.d of 240 V is applied to the primary of a step-down transformer of with 400 turns on the primary and 15 on the secondary. Suppose also that we connect a 5 ohm resistor across the secondary (whose own resistance is much less than 5 ohm). What will be the primary current?

    Step 1: Calculate the secondary voltage using the turns ratio. 9 V agreed?

    Step 2: 9 V will drive a current of 1.8 A through the 5 ohm resistor. Agreed?

    Step 3: If we make the approximation that no energy is dissipated in the transformer itself, then…
    power in = power out, that is [itex]V_p\ I_p = V_s\ I_s[/itex]
    This gives the primary current as 0.0675 A, doesn't it?
     
  13. Apr 29, 2014 #12

    First 2 steps : I agree :smile:

    Third step, why can't we just calculate the current in the primary coil from : Vp / Rp
     
    Last edited: Apr 29, 2014
  14. Apr 29, 2014 #13

    Philip Wood

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    This is a good question. I'm afraid it can't be answered without going quite deeply into transformer theory; more deeply than you may need to do at your present level of study…

    The answer is along these lines… There is a changing magnetic flux in the transformer core, due to the changing current in the primary and, if there is a load connected across the secondary, to the changing current in the secondary. The changing flux induces an emf in the primary, and this emf is usually much larger than the p.d., [itex]I_p R_p[/itex] due to the resistance of the primary coil. So, to a reasonable approximation, we can forget all about the resistance of the primary coil. The primary current is determined by the constraint that the voltage induced in it is equal to the voltage we apply across it. If we pursue this line of argument mathematically, the primary current comes out to be [itex]\frac {n_s}{n_p} \times I_s[/itex] to a good approximation.

    At your present level of study, step 3 is probably all you need. It is perfectly valid, even though it doesn't tell you what's going on inside the transformer.
     
    Last edited: Apr 29, 2014
  15. Apr 29, 2014 #14

    Drakkith

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    Because the resistance (or impedance to be more accurate) of the primary changes depending on the power draw of the secondary. If you cut the impedance of the secondary in half, you double your current and your power draw, which means that the current and power in the primary circuit also doubles. This requires that the impedance be cut in half for the primary circuit as well.
     
  16. Apr 29, 2014 #15

    russ_watters

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    With due deference to Philip: the harder you push on electrons going through a coil - the more you try to push through, the harder they push back.
     
  17. Apr 30, 2014 #16
  18. Apr 30, 2014 #17

    Philip Wood

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    I'm not sure how we can help without knowing where your difficulties lie.

    I think Drakkith is essentially spelling out the implications of Step 3 in my post 11, though Drakkith may disagree. [The resistance (or … impedance) of the primary, referred to by Drakkith is NOT the resistance that you could measure by connecting an ohm-meter across the primary of a transformer not in use. It is an EFFECTIVE resistance or impedance. My advice would be to keep off the idea at this stage in your learning.]

    I can't comment on Russ Water's post.
     
  19. Apr 30, 2014 #18
    Philip wood you are so helpful I really appreciate that :smile:

    I'd like you to help me with the idea of the transformer from the beginning

    We'll talk about 100% ideal transformers first, because ideality makes it easy to understand the real ones after that,

    I've read that the resistance of the coils of an ideal transformer is zero ohm
    If we assume that there isn't a load connected to the secondary what will happen ???
     
  20. Apr 30, 2014 #19

    russ_watters

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    I was trying to convey more concisely why voltage and amperage aren't linearly dependent on each other in a transformer; the more amperage you drive through the primary, the higher the impedance becomes, so it isn't a simple V=ir situation like a resistor is.

    And this is why I was focusing on that:
    Resistors and inductors operate completely different from each other. Zero resistance in a straight wire implies infinite current, but that isn't anywhere close to what happens in an inductor. You'll need to stop thinking of inductors as if they were resistors. A good starting point here would be to read the wiki articles on impedance, inductors and inductance.
     
    Last edited: Apr 30, 2014
  21. Apr 30, 2014 #20

    Drakkith

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    No, I agree. The ohm-meter would measure the simple resistance portion of the impedance, which will be very low for a transformer coil.

    Before you even get into the details of transformers, do you understand what impedance is? Do you know the difference between how a resistor, inductor, and capacitor affect impedance? How about how different frequency AC currents are affected by impedance? Without knowing those things I don't think you'll understand how a transformer works.
     
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