Question

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

Answer 1

Data: L_P = L_S = 2 × 10^-4 H, M = 4 × 10^-6 H
M = K\(\sqrt{L_{\mathrm{P}} L_{\mathrm{S}}}\)
The coupling coefficient is
K = \(\frac{M}{\sqrt{L_{\mathrm{P}} L_{\mathrm{S}}}}=\frac{4 \times 10^{-6}}{\sqrt{\left(2 \times 10^{-4}\right)^{2}}}=\frac{4 \times 10^{-6}}{2 \times 10^{-4}}\)
= 2 × 10^-2
Therefore, the percentage of flux of the primary reaching the secondary is
0.02 × 100% = 2%