- #1

Quadratic64

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## 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.