Matching a 600 Ω amplifier to a 4 Ω speaker using an ideal coupling transformer: what turns ratio should be used for maximum power transfer?

Introduction / Context:Transformers are widely used to match a high-impedance source to a low-impedance load. Correct turns ratio selection ensures the load is reflected as the source's impedance, maximizing power transfer and efficiency.

Given Data / Assumptions:

• Source resistance (amplifier output) Rs = 600 Ω.
• Load (speaker) RL = 4 Ω.
• Ideal transformer and standard matching condition: R_ref at source = Rs.

Concept / Approach:For an ideal transformer, reflected resistance from secondary to primary is R_ref = RL * (Np/Ns)^2. Maximum power transfer requires R_ref = Rs. Express turns ratio as r = Ns/Np per the answer set.

Step-by-Step Solution:Set RL * (Np/Ns)^2 = Rs(Np/Ns)^2 = Rs / RL = 600 / 4 = 150Np/Ns = sqrt(150) ≈ 12.247Hence Ns/Np = 1 / 12.247 ≈ 0.0817Closest option: 0.08

Verification / Alternative check:Using Ns/Np = 0.08 implies Np/Ns ≈ 12.5, giving R_ref ≈ 4 * 12.5^2 = 4 * 156.25 = 625 Ω, close to 600 Ω and consistent with a rounded choice.

Why Other Options Are Wrong:

• 8 or 80: These would be very large Ns/Np and would not reflect 4 Ω up to about 600 Ω.
• 0.8: Too large; would reflect far less than 600 Ω.

Common Pitfalls:Mixing up Ns/Np and Np/Ns or forgetting the square relationship between turns ratio and impedance reflection leads to large errors.

Final Answer:0.08