The greatest power will be dissipated in the resistor if R =?

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SUMMARY

The discussion centers on calculating the resistance (R) in the secondary coil of a transformer, concluding that R equals 4 ohms. The power generated in both the primary and secondary coils is equal, as demonstrated by the equations P1 = N2^2 * 100 and P2 = N1^2 * R. Additionally, the importance of impedance matching in RF circuits is highlighted, emphasizing that maximum power transfer occurs when load impedance matches source impedance.

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Homework Statement
A source with an impedance of 100 Ω is connected to the primary coil of a transformer and a
resistance R is connected to the secondary coil. If the transformer has 500 turns in its primary coil and 100 turns in its secondary coil the greatest power will be dissipated in the resistor if R =?

The answer is 4Ω.
Relevant Equations
N1/N2 = V1/V2 = I2/I1
P = I^2*R
Since the power generated by both primary coil and secondary coil would be the same in a transformer, so I used the relationships stated above to deduce the resistance R in secondary coil.

P1 = N2^2 *100 = 100^2 * 100
P2 = N1^2 *100 = 500^2 * R
R = 4

Let me know if my thoughts here are correct, thanks!
 
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Yep, that's it for the transformer part. Then you add the part about impedance matching. The greatest power delivered is when the load impedance equals the source impedance.

Transformers are often used to help match impedance in RF circuits, just like this. In analysis, you can move an impedance from one transformer winding over to the other if you adjust it's value by N2.
 
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On second thought, your answer is correct, but I'm not convinced that your math matches the words you used to describe it. The more usual approach is to recognize that the load voltage is N times less than the primary voltage and the load current is 1/N times the primary current.
 
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DaveE said:
On second thought, your answer is correct, but I'm not convinced that your math matches the words you used to describe it. The more usual approach is to recognize that the load voltage is N times less than the primary voltage and the load current is 1/N times the primary current.
Thanks for your in-depth analyses!
 
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