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Transformers confusion

  1. Sep 21, 2014 #1
    I know that in a step down transformer the voltage is reduced and therefore current has to increase because P=IV. But potential difference is what drives the current so when the voltage decreases it seems strange for the current not to decrease too. Another why of putting it is if both coils were made out of the same material then wouldn't they have the same resistance? and so if they both obey V=IR then when V goes down and R remains the same then I would need to go up but it doesn't. why?
     
  2. jcsd
  3. Sep 21, 2014 #2

    phinds

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    You have a fundamental misunderstanding. What happens in a step down transformer is that the output potential is stepped down from the input potential. The current in the output is whatever current is required to drive whatever load is supplied. This is, of course, subject to the capabilities of the transformer. Since the output power can be as much as the input power (ignoring inefficiencies in the transformer) that means that if the voltage source feeding the transform can supply a max of A amps, then the current capability of the output of a step down transformer is nA where n is the amount of the step down. That doesn't mean it HAS to supply that much current, just that it can if the load requires it.
     
  4. Sep 21, 2014 #3
    Yes. A more fundamental approach is necessay.

    First off , V=IR, is applicable only for a resistor, where the voltage across the resistor can be considered as the driving force. Since R is considered to mainly constant, changing the voltage will also change the current.

    For other electrical components, such as capacitance, C, or inductance, L, the relationship between V across the component and the current I flowing through the component does not have this direct one-to-one relationship.

    Split you circuit into three seperate areas - the load R with voltage V(r) and I(r) , the transformer, and the supply with V(s) and I(s).
    At the load, V(r) = I(r) R. With that, using the transformer ratio, and knowing the supply voltage V(s), you can determine the supply current I(s).
     
  5. Sep 21, 2014 #4

    Drakkith

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    Let's look at a very simplified example. Consider a transformer with 100 volts on the primary side, and 10 volts on the secondary side, for a step down ration of 10:1 and a turn ration of 10:1.
    Since the primary circuit consists only of the voltage supply and the primary coil, the resistance of the primary side never changes. For this example let's assume that the resistance of both the primary coil and secondary coil are zero.

    Now, let's first look at the secondary side as if it were a DC circuit with a variable resistor capable of 1 ohm, 10 ohms, and 100 ohms (remember that the resistance of the secondary coil is zero in our example). Starting the with 100 ohm resistor, we see that the voltage is stepped down from 100 to 10 volts, so the current in the secondary is 10/100 = 0.1 amps. Power in the secondary side is thus 0.1x10, or 1 watt.

    Changing the resistor to 10 ohms gives us 10v/10 ohms = 1 amp. 1 amp x 10 volts = 10 watts of power in the secondary circuit.

    Changing the resistor to 1 ohm: 10v/1 ohm = 10 amps. 10 amps x 10 v = 100 watts of power circuit.

    Notice that we have only talked about the power of the secondary circuit, not the primary. To understand the primary we have to understand that resistance is actually a specific subset of a broader term called impedance. Impedance includes resistance along with both inductive reactance and capacitive reactance and it is crucial to understand impedance when discussing alternating current and transformers. I can't teach everything about impedance here, so I'll just keep it simple.

    In our simplified transformer example we can worry only about resistance and inductance and ignore capacitance. Now, understand that inductance is the property of a conductor whereby a change in current induces a voltage in both the conductor itself and other nearby conductors. In contrast, inductive reactance is the resistance the circuit has to the change in the flow of current. In other words, inductive reactance acts just like plain old electrical resistance for an AC circuit.

    Since a transformer is necessarily an AC device (except in very specific cases which we won't talk about here), and we've already discussed the normal resistance portion of a circuit in the above part of the post, we only need to discuss the inductive reactance. For the sake of discussion let's assume that the inductive reactance is 100 ohms and the capacitive reactance to be zero (to simplify things). Since electrical impedance is total reactance + resistance, and total reactance is the inductive reactance minus the capacitive reactance, our total impedance is going to be the resistance of the resistor + inductive reactance in this example.

    With 100 ohm resistor:
    The maximum current in the secondary side with this load would be 0.1 amps if we just had a DC circuit. However, since this is an AC circuit we have use impedance instead. Adding 100 ohms from the resistor to the 100 ohms from the inductive reactance gives us 100+100 = 200 ohms of impedance. So the average current through the circuit is: 10v/200 ohms = 0.05 amps. Note that this current fluctuates in both magnitude and direction, but the average current is 0.05 amps. Power: 0.05 amps x 10 volts = 0.5 watts.

    Because the current is fluctuating, it is generating a changing magnetic field that cuts the windings of both the secondary and primary coils. The end result is that the changing flux from both the primary and secondary coils causes the primary and secondary circuits to have equal power. So, since our secondary current is 0.05 amps, and our voltage is 10 volts, the power in the secondary is 0.05 x 10 = 0.5 watts. Since the primary power has to equal the secondary, we can use it to find the impedance in the primary. We have V = 100 volts, so the power is 100v x I = 0.5 watts, and the current is: I = 0.5 / 100 = 0.005 amps. Finding impedance: 0.005 amps = 100v/impedance, or 100v/0.005 amps = Impedance. So impedance is 10/0.005 = 20,000 ohms.

    So, the primary side: 100 volts x 0.005 amps = 0.5 watts.
    Secondary side: 10 volts x 0.05 amps = 0.5 watts.
    Power is equal in both cases.
    Impedance is 20,000 ohms on the primary side and 200 ohms on the secondary side.

    With the 10 ohm resistor:
    Secondary:
    Impedance is 110 ohms, so: 10v/110 ohms = 0.09 amps.
    Power is 0.09 x 10v = 0.9 watts on the secondary side.

    Primary:
    Current: 0.9w watts = 100v x I, or 0.9/100v = I. So I = 0.009 amps.
    Impedance: 0.009 amps = 100v/Impedance, or 100v/0.009 amps = impedance, or 100 / 0.009 = 11111 ohms.
    Checking math: 100v/11111 ohms = .009 amps. 100v x 0.009 amps = 0.9 watts.


    With 1 ohm resistor:
    Secondary:
    Impedance is 101 ohms, so: 10v/101 ohms = 0.099 amps.
    Power: 0.099 x 10v = 0.99 watts.

    Primary:
    Power is 0.99 watts.
    Current: 0.99 watts = 100v x I, or 0.99/100 = 0.0099 amps.
    Impedance: 0.0099 = 100/Impedance. So, 0.0099 x Impedance = 100, or Impedance = 100v/0.0099 amps = 10101 ohms.

    Notice the pattern of the impedence between the primary and secondary sides for each example? The ratio of impedance between each side is equal to the turn ratio and voltage ratio. So as you change the impedance on the secondary side by connecting or disconnecting loads, the impedance on the primary side fluctuates as well to keep the power in both sides equal. This also requires that the current go up when the voltage is stepped down in order to keep the power between each side equal, as you can see in the above examples.

    In addition, as either the resistance or the inductive reactance portions of the impedance of the secondary side changes, only the inductive reactance of the primary coil changes in response since the resistance portion of the impedance stays the same.

    I hope all that makes sense.

    Note: This was a simplified example and may not take into account all of the complexities of an AC circuit. I trust that the overall idea is correct, but some of the details and math may be slightly off. As always, someone correct me if I'm wrong.
     
    Last edited: Sep 22, 2014
  6. Sep 21, 2014 #5

    davenn

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    nice post Drak :)
     
  7. Sep 21, 2014 #6

    phinds

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    I absolutely agree, but I'm thinking he has WAY too much spare time on his hands and needs to get out more :D
     
  8. Sep 21, 2014 #7

    davenn

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    hahaha and his post count confirms that theory ;)
     
  9. Sep 22, 2014 #8

    Drakkith

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    Heh, that post looks way longer after it was posted than when I was typing it. :rolleyes:
     
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