Does the load on a transformer affect its efficiency?

[Quadratic64]

Homework Statement

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.

Homework Equations

V=IR
η=P in / P out
P=IV
Np/Ns=Vp/Vs

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.

 

[voko]

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.

 

[Quadratic64]

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?

 

[voko]

Correct.

 

[Nickie]

The load on a transformer can affect its efficiency, but it is not the only factor. The efficiency of a transformer is determined by the energy losses in the windings and core, as well as the load connected to it. In this case, the efficiency of the transformer is 50%, which means that 50% of the input power is being lost in the transformer itself. When you change the load from a 50Ω lamp to a 100Ω lamp, the current and voltage on the secondary side change, resulting in a different power output. This change in power output affects the overall efficiency of the transformer. It is important to note that the efficiency of a transformer can also be affected by the type of load connected to it, as well as the power factor of the load. In conclusion, the efficiency of a transformer is not solely based on its own characteristics, but also on the load connected to it.