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A question on a transformer?

Discussion in 'Electronic Basics' started by henry chow, Nov 15, 2005.

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  1. henry chow

    henry chow Guest

    Last time I posted a question about power lost on the transmission line and
    I thank everyone for answering my question. Right now I get a better
    understanding now because the V (voltage) (used on the power calculation) is
    measured across the single line rather than between the two high-voltage
    lines which is over ten thousand volts.
    Here is another "basic" question on a transformer. Suppose I have a
    transformer with a primary windings of 100 turns and a secondary windings of
    200 turns. If I apply an AC voltage of 100 volts and 2 A to the primary, I
    will get an AC voltage of 200 volts and 1A from the secondary.
    If I put a load of, say 50 ohms, across the secondary windings output, what
    should the current be? According to Ohms' law, it should be I=V/R=200
    volts/50 ohms=4 A. But the max current you can get from the secondary
    windings is 1A. Why the contradiction? If the current across the load is
    1A, should the voltage drop across the load be V=IR=1A * 50 ohms=50 volts,
    which is not right because the secondary windings have a voltage of 200
    volts? Can you explain the contradiction?

    Again, suppose instead I apply a load of 400 ohms to the secondary windings,
    I should get a current of 200 volts/400 ohms=0.5 A? Is that correct? Then
    what happens to the 1A I am supposed to get from the secondary windings?
     
  2. Bob Myers

    Bob Myers Guest

    ....IF the load connected to the secondary happens to be 200 ohms;
    otherwise, the current through the secondary (and therefore, the current
    in the primary, transformed from the secondary current by the turns
    ratio) will be whatever the load requires it to be.

    This is a basic error commonly made by beginners in just about every
    situation in electronics - normally, you CAN'T simply place X volts
    across and Y amps through a given pair of terminals, since there generally
    isn't any such thing as a load that will alter itself to permit that. You
    put
    X volts across a load, and the current will be determined by the load
    impedance; you put Y amps THROUGH a load, and similarly the
    voltage across it is determined by the load impedance. You can't (at
    least at this simple level) get around Ohm's Law.

    Consider what you see at the primary of a transformer when NO load
    is connected to the secondary - better yet, let's assume the transformer
    is in a "black box" so that you don't even see that it IS a transformer -
    all you have access to are the primary terminals. The secondary
    terminals, hidden inside the box, aren't connected to anything. Now,
    you walk up with, say, a 100 VAC source, and you connect it to the
    terminals you see on this "black box." What happens?

    Another aspect of transformers to be aware of - don't confuse the
    secondary current RATING with what current you will get out of
    the transformer in any given circumstance. If you see a transformer
    which has specs something like, say, "Primary: 120 VAC 1 A;
    Secondary: 6.3 VAC 12A" this does NOT mean that the primary
    will always draw 1 A or the secondary will always produce 12 A
    out. What it means is that you can hook up 120 VAC to the primary
    and expect to get 6.3 VAC from the secondary, and that you had
    better not arrange your input voltage and the load on the secondary
    so that the primary current exceeds 1A or the secondary current
    exceeds 12A - 'cause that's all those windings are rated to take.

    Bob M.
     
  3. henry chow wrote:
    (snip)
    You apply voltage across the primary, and the transformer passes as
    much current as it passes. In the above example, the primary current
    would be 2 amps only if the secondary load current were 1 amp and it
    was an ideal transformer (no losses, perfect magnetic coupling between
    the windings, and no magnetizing current).
    If the transformer were large enough to supply that current without
    sagging its output voltage.
    There is no contradiction because there is no way to specify the
    primary current and the primary voltage, independent of the secondary
    load. You can control the primary current to be 2 amps, but that
    requires you to change the primary voltage to whatever produces that 2
    amp current. If you did that, with the 50 ohm resistor connected to
    the secondary, the resistor current would be 1 amp, but that is
    because the secondary voltage would have to have fallen to 50 volts,
    which means that the only way the primary current could be 2 amps
    under these conditions would be if you lowered the primary voltage to
    25 volts. Then, the voltage and current ratios would remain consistent.
    I hope I have.
    It falls to 0.5 amp, and the primary current falls to 1 amp. The load
    resistor reflects back through the transformer to the primary voltage.
     
  4. Pooh Bear

    Pooh Bear Guest

    You can't do that. The current will follow the load placed on the transfomer.
    There will be a magnetising current though due to the primary inductance.

    Using the very simple model, power in = power out. That will answer all your
    questions.

    Real transformers have losses though ( so power out < power in ).

    Graham
     
  5. Jasen Betts

    Jasen Betts Guest

    you get 4A until the transformer overheats and fails. or until a fuse
    somewhere blows. while this is happening there is 8A going through the
    primary.
    yes, and the current in the primary would be 1A - well slightly more because
    transformers need a /magnetising/ current to make them work.
     
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