Calculate the equivalent inductance of the following inductive circuit.

Applying the formula for inductances in series and parallel will allow us to determine the equivalent inductances of each circuit one at a time. For any more questions, see the answer below.

The mix of series and parallel inductances makes up the provided circuit. After determining the equivalent inductance for each component, we started from the right side and drew equivalent circuits.

Here, ${\mathrm{L}}_{67}={\mathrm{L}}_6+{\mathrm{L}}_7=40+100=140\mathrm{mH}$ as (L6 and L7 are in series )
${\mathrm{L}}_{567}=\frac{1}{1/{\mathrm{L}}_5+1/{\mathrm{L}}_{67}}=\frac{1}{1/50+1/140}=36.84\mathrm{mH}$ as (L5 and $\mathrm{L}67$ are in parallel)
${\mathrm{L}}_{4567}={\mathrm{L}}_4+{\mathrm{L}}_{567}=20+36.84=56.84\mathrm{mH}$
${\mathrm{L}}_{34567}=\frac{1}{1/{\mathrm{L}}_3+1/{\mathrm{L}}_{4567}}=\frac{1}{1/30+1/56.84}=19.64\mathrm{mH}$
${\mathrm{L}}_{234567}={\mathrm{L}}_2+{\mathrm{L}}_{34567}=40+19.64=59.64\mathrm{mH}$
$\text{ So, }{\mathrm{L}}_{\mathrm{EQ}}={\mathrm{L}}_{1234567}=\frac{1}{{ \mathrm{L}}_1+1/{\mathrm{L}}_{234567}}=\frac{1}{1/20+1/59.64}=14.98~15\mathrm{mH}$

Hence, the correct answer is option C.