Coupling effects: A transformer has a turns ratio of 1:1 and a coupling coefficient k = 0.85. If 2 V AC is applied to the primary, what RMS voltage appears at the secondary?

Introduction / Context:
Real transformers do not couple all magnetic flux perfectly. The coupling coefficient 0 ≤ k ≤ 1 captures how effectively primary flux links the secondary. This question tests applying k to estimate secondary voltage when the ideal turns ratio alone would suggest equality.

Given Data / Assumptions:

• Turns ratio Np:Ns = 1:1.
• Coupling coefficient k = 0.85.
• Primary RMS voltage Vp = 2 V.
• Assume negligible resistance and leakage beyond k's effect on induced voltage.

Concept / Approach:
With imperfect coupling, the induced secondary voltage magnitude is approximately Vs ≈ k * (Ns/Np) * Vp. With Ns/Np = 1, Vs ≈ k * Vp = 0.85 * 2 V.

Step-by-Step Solution:

Verification / Alternative check:
In the ideal case (k = 1), Vs would be 2 V. Reducing k to 0.85 scales the induced magnitude to 85% of ideal, consistent with 1.7 V.

Why Other Options Are Wrong:

• 0.85 V: Would correspond to applying k twice or using Vp = 1 V.
• 1 V: Not supported by the given k and Vp.
• 0 V: Only true for DC excitation or open primary; not here.
• 2 V: Would require perfect coupling (k = 1).

Common Pitfalls:

• Ignoring k and assuming ideal behavior.
• Confusing k with efficiency; k relates to flux linkage, not directly to power loss.

Final Answer:
1.7 V