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Mutual Inductance and Transformer Question

  1. Oct 11, 2009 #1
    Hi, I would like to ask a question about Mutual Inductance and Transformer.

    Suppose I have two coils N1 and N2, where N1 is the primary coil and N2 is the secondary coil that is connected in series with some resistance Rx. If I place a voltage V1 across the primary coil, the output voltage V2 across the secondary coil is N2/N1*V1. Also, if a current A1 is flowing in the primary coil, then the current A2 flowing in the secondary coil is V1/V2*A1.

    Now, if I connect a resistor Ry in parallel with Rx (or do anything to reduce the resistance), the overall resistance of the secondary coil decreases, and therefore A2 increases. And since the voltages did not change (anything you do to the secondary coil is not supposed to affect the voltage of the primary coil), A1 should also increase.

    Now this is where my confusion is. How can the voltage of the primary coil remain the same when the current increases? By Ohms law, shouldn't the voltage of the primary coil also increase?

    Thanks in advance.
     
  2. jcsd
  3. Oct 11, 2009 #2

    sophiecentaur

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    The resistance measured across the primary has reduced because you added the second resistor. With no resistor across the secondary (I.e. Zero conductance) the input resistance is infinite (ideally).
     
  4. Oct 11, 2009 #3
    Hi Simpleton.

    You have to consider the transformer as a coupled system. Changes in the primary circuit affect the secondary as you have said, but the opposite is also true. Changes in the secondary circuit are 'reflected' to the primary.

    If you poke at the relationships between primary and secondary voltage and current, you will find that the impedances are related by the square of the turns ratio.

    You can then calculate the impedance 'seen' at the primary terminals due to a load at the secondary terminals. Try it for an open circuit, a short circuit and for an arbitrary resistance.
     
  5. Oct 13, 2009 #4
    Hi simpleton-
    A change in the current of the primary winding induces a voltage in the secondary winding. For ac voltages and currents, this induced voltage is 90 degrees out of phase with the primary current. This based on the Faraday Law of induction. The current in the secondary winding is dependent on the load (resistor, etc.) attached to the secondary.
    Bob S
     
  6. Oct 14, 2009 #5
    Hi sophiecentaur,

    Can you explain how the resistance measured across the primary has been reduced? I don't know how you incorporate the secondary transformer into the equation and how it reduces the resistance.

    And to Pogo and Bob S, yes I can understand that changes to the secondary transformer affects the primary one, in the sense that if the resistance is reduced, then the secondary circuit would draw a bigger power, so the current in the primary circuit would increase. However, that still does not explain why the voltage did not increase. I just think that the voltage should increase together with the current, but then the voltage is supposed to be increase too, shouldn't it? I cannot see how ohm's law and the transformer law can gel together.

    Can someone show me how you take into account the resistance by the secondary circuit in a kirchoff's loop rule of the primary transformer or something? How do all these voltages sum to 0?

    By the way, I mean the change in root mean square voltage and current: Imax/sqrt(2) and Vmax/sqrt(2) when I am talking about the increase in current and voltage. I guess that is the value you use in your calculations rit?

    Thanks.
     
  7. Oct 14, 2009 #6

    sophiecentaur

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    Simpleton
    The first question I could ask you is how could the voltage across the primary increase if it is connected to a low impedance (mains, for instance) ac power supply?
    It just HAS to stay at the supply voltage.
    Then, the increased current in the secondary circuit will induce an appropriate voltage in the primary so that more current flows - presenting the supply with a lower resistance.

    I don't think there's a more straightforward explanation short of deriving the transformer equations, which is do-able but hard work. I'm sure Wikkers has it.
     
  8. Oct 14, 2009 #7
    Actually, the part I am confused about is exactly this point. As the primary transformer is connected to a constant source of ac, how can it change its voltage? However, if its current increased, how can its voltage not change? And that was where I got really confused.

    If I did not get you wrong, are you trying to say that the primary tranformer induces the secondary transformer, and the secondary transformer will induce the primary transformer and this will somehow reduce the resistance and thus increase the current?

    However, I am still not very comfortable with this. I guess that you also use the simple transformer law to find out the induced voltage by the secondary transformer on the primary transformer. However, if that is the case, even if I change the resistance of the secondary transformer so that there is a larger current, the voltage induced in the secondary transformer is still the same. So the induced voltage on the primary transformer by the secondary transformer is still the same rit? So there should be no difference.

    What is wrong with my reasoning? Thanks in advance.
     
  9. Oct 14, 2009 #8

    sophiecentaur

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    Would you not agree that a perfect transformer with no load on the secondary will present an infinite resistance at the primary input?
    Any current which flows through the secondary will result in a current in the primary (I.e. looks like a resistance). This has to be because of magnetic coupling between the two windings. Talking in terms of induced voltages due to currents and taking into account the phases gives 'the right answer' in the form of the transformer equations. With no load, the self inductance of the primary produces a back emf equal to the supply volts (assuming there is enough self inductance). Any secondary current induces a further emf which reduces this back emf and allows current to flow.
    Rather than trusting in your intuition it may be better to follow the actual sums through. It is just a bit of 'bookwork' which is self consistent and just works.
     
  10. Oct 15, 2009 #9

    sophiecentaur

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    I have re read your last post, simpleton, and I may have spotted something which could help.
    The induced voltage in the secondary will be proportional to the turns ratio, however much current flows and will lag by pi/2. BUT the induced emf in the primary will be due to the Current (actually, the rate of change of current - so you get the time differential) in the secondary times the mutual inductance and another pi/2 phase change, giving a voltage opposite to the back emf. More current - more emf subtracted from the back emf (which is due to the primary self inductance)- allowing more current to flow into the primary from the source (=lower input resistance).

    To calculate the no-load volts situation for a transformer you work out the field due to the primary (proportional to the number of primary turns) and then the induced emf per turn is given by the rate of change of flux, so the total output emf will be proportional to the total number of secondary turns. That's where the simple transformer equation comes from. (I think that you need to assume some current flows in the primary for this, so even an ideal transformer has to have a finite admittance.)
     
  11. Oct 15, 2009 #10
    The perfect transformer will have some primary inductance, so there will be a small current. even when the transformer has no load.

    INL = Vpri/jwL

    where j is a 90 degree phase shift, and w = 2 pi F (2 pi times frequency). So the primary NL (no load) current is 90 degrees out of phase with the voltage due to the primary self inductance. But when you attach the transformer secondary to a resistive load, there is an additional primary current that is in phase with the voltage.
    Bob S
     
  12. Oct 16, 2009 #11

    sophiecentaur

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    Bob S
    The perfect transformer will have (near) Infinite Primary inductance (near zero Admittance). No current must pass when there is no load.
    That formula is, of course, right. I was trying to describe in words why no current flows in terms of back emf - a familiar, cuddly, concept. It is true to say that this back emf isn't in exact antiphase with the supply - hence some primary current flows.

    I don't think there is actually a satisfactory arm waving description of how a transformer works but I try.
     
  13. Oct 17, 2009 #12
    Hi sophie-
    Most transformer "theory" papers on the web don't discuss the reactive phase shifts in single phase transformers. This one does:
    http://sound.westhost.com/xfmr.htm
    Bob S
     
    Last edited by a moderator: Apr 24, 2017
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