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

_{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%