Given, Series equivalent inductance, Leq=8H Parallel equivalent inductance, Leq2=1.5H=
3
2
H Let two inductors used here be L1 and L2, then Leq1( series )=L1+L2 ∴Leq1=L1+L2=8 ...(i) and
1
Leq1
(parallel) =
1
L1
+
1
L2
⇒Leq2=
L1L2
L2+L1
⇒
3
2
=
L1(8−L1)
8
⇒12=8L1−L12 ⇒L12−8L1+12=0 ⇒L12−6L1−2L1+12=0 ⇒L1(L1−6)−2(L1−6)=0 ∴L1=2H or 6H and L2=(8−2) or (8−6) =6H or 2H ∴ Difference between inductances is (L1−L2) or (L2−L1)=6−2 = 4H