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 = $\frac{M}{\sqrt{L_{\mathrm{P}}{L}_{\mathrm{S}}}}=\frac{4\times {10}^{-6}}{\sqrt{{(2\times {10}^{-4})}^2}}=\frac{4\times {10}^{-6}}{2\times {10}^{-4}}$
= 2 × 10^-2
Therefore, the percentage of flux of the primary reaching the secondary is
0.02 × 100% = 2%