The primary and secondary coil of a transformer each have an inductance of 200 `xx10^(-6)`H. The mutual inductance (M) between the windings is 4 `xx 10^(-6)`H. What percentage of the flux from one coil reaches the other?

Answer

To find the percentage of the flux from one coil that reaches the other in a transformer, we can use the concept of the coefficient of coupling (k). The coefficient of coupling is defined as the ratio of the mutual inductance (M) to the geometric mean of the inductances of the primary (L1) and secondary (L2) coils.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Inductance of primary coil (L1) = 200 × 10^(-6) H
- Inductance of secondary coil (L2) = 200 × 10^(-6) H
- Mutual inductance (M) = 4 × 10^(-6) H

2. **Calculate the Geometric Mean of Inductances:**
\[ \text{Geometric Mean} = \sqrt{L1 \times L2} = \sqrt{(200 \times 10^{-6}) \times (200 \times 10^{-6})} \]
\[ = \sqrt{(200^2) \times (10^{-6})^2} = \sqrt{40000 \times 10^{-12}} = 200 \times 10^{-6} \text{ H} \]

3. **Calculate the Coefficient of Coupling (k):**
\[ k = \frac{M}{\sqrt{L1 \times L2}} = \frac{4 \times 10^{-6}}{200 \times 10^{-6}} \]
\[ = \frac{4}{200} = 0.02 \]

4. **Convert Coefficient of Coupling to Percentage:**
\[ \text{Percentage of flux} = k \times 100 = 0.02 \times 100 = 2\% \]

### Final Answer:
The percentage of the flux from one coil that reaches the other is **2%**.