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Homework Help: Does the load on a transformer affect its efficiency?

  1. Apr 11, 2013 #1
    1. The problem statement, all variables and given/known data

    A transformer has 400 primary coils and 200 secondary coils. The primary voltage is 200 V, and the current is 2 A. A lamp is connected to the secondary coil. The efficiency of this transformer is 50%. Find the resistance of the lamp.

    2. Relevant equations

    η=P in / P out

    3. The attempt at a solution

    From the turns ratio and the primary voltage, I found that the secondary voltage is 100 V. Since the efficiency is 50% and the input power is 200V*2A=400W, the output power is 50%*400W=200W. The current is 200W/100V=2 A.

    Now I have the current and voltage on the secondary side, so I should be able to use V=IR to find the resistance of the lamp. So R=V/I=100V/2A=50Ω.

    4. My question

    But I also noticed something strange about this problem. What happens if I replace the 50Ω lamp with a 100Ω lamp? The induced voltage in the secondary coil is the same (100V), but using V=IR, I get that the current is 100V/100Ω=1 A. Therefore, the power is 1A*100V=100W, and the efficiency of the transformer must be η=100W/400W=25%.

    My conclusion seems very counter-intuitive, since I don't understand how changing the load can change the efficiency of a transformer. The efficiency should only be based on the characteristics of the transformer itself (the energy lost in the windings and the core), not the devices that I connect to it. Is there something wrong with the problem, or was the way I solved it incorrect? I can't find much information about efficiency on the Internet, so any help will be appreciated. Thanks.
  2. jcsd
  3. Apr 11, 2013 #2
    Your mistake is in assuming that no matter how much power is used in the secondary, the power in the primary is constant. Using this logic, you can easily get the power in the secondary greater than in the primary, which is manifestly impossible.
  4. Apr 11, 2013 #3
    So when I connect the 100Ω lamp, the secondary power is 100W. Then the primary power Pp=Ps/η=100W/50%=200W. Therefore, the current on the primary side will be reduced to I=P/V=200W/200V=1 A. Am I correct?
  5. Apr 11, 2013 #4
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