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What is it that causes electric current to drop after a step up ?

  1. Mar 6, 2014 #1
    what is that causes the current in the wire that has a stepped up voltage * in the transformer , the wire with more coils * to have a lower current ?
     
    Last edited: Mar 6, 2014
  2. jcsd
  3. Mar 6, 2014 #2

    SteamKing

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    Are you talking about a transformer?
     
  4. Mar 6, 2014 #3
    yes , after stepping up the voltage
     
  5. Mar 6, 2014 #4

    Dale

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    Impedance. Or equivalently the conservation of energy.
     
  6. Mar 7, 2014 #5

    davenn

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    and to elaborate on what dalespam said

    power out = power in ( minus a some losses in the transformer )

    Idealy, so if the power in on the primary side is 200V @ 1 Amp = 200W ---- power out on the secondary cannot be more that 200W and as stated it will be a little less.
    So if you are stepping up from 200V to 400V ( ie double the voltage) you are going to halve the available current
    so idealy, you will have 400V @ 0.5A = 200W
    in the real world, you will have a little less than 0.5A

    cheers
    Dave
     
  7. Mar 7, 2014 #6
    yea i totally understand , but i mean , physics works in order to conserve energy , so there must be some mechanism that does its job such that power is always conserved , what is this mechanism exactly , impedance ?
     
  8. Mar 7, 2014 #7
    Ampere's Law, AL. To establish core flux, magnetizing current exists in the primary. The primary is powered by a CVS (constant voltage source). Flux is determined by voltage, frequency, core area, and number of primary turns Np. Open circuited secondary current Is = 0, primary current Ip = exciting current (magnetizing current plus hysteresis/eddy current loss).

    When loaded on secondary, Is becomes non-zero and generates a flux since current in a winding has a flux. This flux, per LL (law of Lenz) is of opposite polarity to the original. The core flux drops due to partial cancellation. The primary unloaded had a counter-emf nearly equal to source emf, the difference appearing across the primary winding impedance. Loaded, the counter-emf drops, so that the primary winding impedance has a voltage equal to the source emf minus a reduced cemf, resulting in larger primary current Ip. This Ip has a field and brings the flux back close to its original value so that secondary voltage is almost equal to the unloaded value.

    Ampere's law AL, states that balance is achieved when the amp-turns of the primary balance that of the secondary, i.e. NpIp = NsIs. The transformer is self regulating, errors incurred since windings have resistance and leakage inductance. So the flux and voltages do not quite reach their original unloaded value, but come within a couple percent.

    Claude
     
  9. Mar 7, 2014 #8

    sophiecentaur

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    I suggest that you are putting things the wrong way round. The output voltage of a step up transformer is higher than the input. The current that flows in the secondary is determined by the value of the secondary Load. So this consequential current is what counts in working out the current that will flow in the primary by affecting the emf that is caused in the primary and, hence, the apparent load that the supply sees.

    The chain of causes and effects is:
    Input volts - [turns ratio] - output volts - [ load] - secondary current - [turns ratio] - primary current
     
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