Two coils inductance $1 \mathrm{H}$ and $3 \mathrm{H}$ are connected series. Their mutual inductance is $5…
Two coils inductance $1 \mathrm{H}$ and $3 \mathrm{H}$ are connected series. Their mutual inductance is $5 \mathrm{H}$. The self-inductance of the combination is
$10 \mathrm{H}$
$28 \mathrm{H}$
$14 \mathrm{H}$
$40 \mathrm{H}$
Solution
For inductors in series:
$\mathrm{L}=\mathrm{L}_1+\mathrm{L}_2+2 \mathrm{M}=(1+3+2 \times 5) \mathrm{H}=14 \mathrm{H}$