Coefficient of coupling (k): If 4% of the primary flux does not link the secondary, what is the coupling coefficient of the transformer?

Introduction / Context:
The coefficient of coupling, k, measures how effectively magnetic flux produced by a primary winding links to the secondary winding. It ranges from 0 (no coupling) to 1 (perfect coupling). Practical transformers seek high k for efficient energy transfer.

Given Data / Assumptions:

• Flux not linked to secondary = 4% of total primary flux.
• Thus, flux that does link = 96% of total.
• Assume standard definition: k equals the fraction of total primary flux linking the secondary in magnitude.

Concept / Approach:
By definition, k ≈ linked flux / total primary flux for tightly coupled windings. If 4% is lost (leakage flux), 96% couples into the secondary.

Step-by-Step Solution:
Leakage flux = 4% ⇒ Coupled fraction = 100% - 4% = 96%k = 0.96

Verification / Alternative check:
A well-designed transformer typically has k close to 1 but not exactly 1 due to leakage inductance; 0.96 is realistic and consistent with the statement of 4% uncoupled flux.

Why Other Options Are Wrong:

• 0.4 and 9.6: Outside the valid 0–1 range for k (9.6 is impossible; 0.4 would imply 60% leakage).
• 4: Dimensionally and numerically incorrect; k must be ≤ 1.

Common Pitfalls:
Confusing percent with fractional values (e.g., taking k = 96 instead of 0.96) or forgetting that k cannot exceed 1.

Final Answer:
0.96