Transformer reflected impedance: A 200 Ω load is connected across the secondary of a transformer whose turns ratio is 4. What value of load does the source 'see' reflected to the primary side?

Introduction / Context:
This problem checks understanding of how a transformer reflects impedances from one side to the other. Being able to compute the reflected resistance is crucial when matching sources and loads for maximum power transfer and efficient design in power electronics and audio systems.

Given Data / Assumptions:

• Load on secondary, RL(sec) = 200 Ω.
• Turns ratio is stated as 4 (interpreted here as Ns/Np = 4, a common convention in basic problems).
• Ideal transformer behavior; no winding resistance or core loss.

Concept / Approach:
The impedance reflection rule for an ideal transformer is: R_ref(primary) = RL(sec) * (Np/Ns)^2. If the given turns ratio is Ns/Np, then Np/Ns = 1 / (Ns/Np). A larger number of secondary turns steps the voltage up and the impedance down on the primary side by the square of the turns ratio inversion.

Step-by-Step Solution:
Given Ns/Np = 4 ⇒ Np/Ns = 1/4R_ref = RL * (Np/Ns)^2R_ref = 200 * (1/4)^2R_ref = 200 * 1/16 = 12.5 Ω

Verification / Alternative check:
If the transformer steps voltage up by 4, it steps current down by 4, and thus impedance is divided by 16 when seen from the primary. 200 Ω / 16 = 12.5 Ω, confirming the calculation.

Why Other Options Are Wrong:

• 50 Ω: Only divides by 4, not by 16.
• 800 Ω: Would be the reflection if Np/Ns = 4, not Ns/Np = 4.
• 0 Ω: An ideal short is not created here.

Common Pitfalls:
Misinterpreting the turns ratio direction or forgetting that impedance reflection scales with the square of the turns ratio rather than the ratio itself.

Final Answer:
12.5 Ω