AP EAPCET 19-Aug-2021 Shift 1 Paper

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