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DSSSB JE EE 2019 Official Paper (Held on 25 Oct 2019)

Option 4 : 87.5 mH

__Concept:__

**The equivalent inductance of series aiding connection is**

L = L1 + L2 + 2M

**The equivalent inductance of series opposing connection is**

L = L1 + L2 – 2M

**Equivalent inductance of parallel aiding connection is**

L = \(\frac{{{L_1}{L_2} - {M^2}}}{{{L_1} + {L_2} - 2M}}\)

**Equivalent inductance of parallel opposing connection is**

L = \(\frac{{{L_1}{L_2} - {M^2}}}{{{L_1} + {L_2} + 2M}}\)

__Calculation:__

Given,

L_{1} = 100 mH

L_{2} = 200 mH

M = 50 mH

Inductance of parallel aiding connection is

\(L=\frac{{{L_1}{L_2} - {M^2}}}{{{L_1} + {L_2} - 2M}}=\frac{100\times 200-50^2}{100+200-2(50)}=87.5\ mH\)