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Homework Help: Transformers Coursework- Interpretation?

  1. Mar 26, 2009 #1
    did an investigation on Transformers- something I thought would be a good idea. The aim was to find out how the amount of resistance put onto the transformer affects the power loss. By having a step-up transformer 300 turns on the primary and 600 turns on the secondary, then by connecting a resistor to the secondary coil (this was the independent variable and was changed: 1, 2, 5, 10, and 200 Ohms). I measured the Voltage and current on the primary coil, and the voltage and current of the secondary coil.

    The following were the recorded measurements (the units are Ohms, Volts, Amperes):

    Resistor Ohm| Primary volt| Primary current| Secondary volt| Secondary current
    5 | 5.43 | 0.43| 1.16| 0.2
    10 |5.42 |0.42| 2.24| 0.2
    1 |5.42 |0.43| 0.27| 0.21
    2 |5.44 |0.44| 0.5| 0.21
    200 |5.42 |0.08| 10.15| 0.03

    So I calculated the Power in, Power Out- and therefore the amount of power in percentage still present after the transformation
    P in P out Pin/Pout *100
    2.3349 0.232 9.936185704
    2.2764 0.448 19.6801968
    2.3306 0.0567 2.43284991
    2.3936 0.105 4.386697861
    0.4336 0.3045 70.22601476

    My question is how do I interpret these results.... (as now I am in doubt as if they make an y sense at all). As theoretically the current should decrease, and the voltage should increase in an up-step transformer. However the results suggest that as the resistance on the secondary coil is increased...the voltage increases..however the current stays the same (except at 200 Ohm resistance). I tried looking for some theory and equations on this topic but couldnt find anything. The results also suggest that as the resistance is increased the power loss is decreased. So if anyone could give me any pointers, explanations or jsut any input.

    By the way when the secondary voltage was measured with no resitor connected it came out to be: 10.46V so about double of the input voltage of about 5V.
  2. jcsd
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