Electrical Circuits EEE In terms of the applied voltage V , the individual Inductances L1 ,L2, … Ln the above equation
can be written as:
i = (1/L1
vdt +(1/L2)
vdt + … +(1/Ln)
vdt = [(1/L1) +(1/L2) + … +(1/Ln )]
vdt.....(2)
Similarly for the circuit in fig(b) we can write the governing equation as:
i = (1/Leq )
v....(3)
where Leq is the equivalent Inductance of all the Inductances L1 , L2 ,….and Ln in parallel.
Since current is the same in the above two equations we find that
1/ Leq = ( 1/ L1 + 1/ L2 + ….. 1/ Ln )
Hence Leq of a parallel circuit consisting of n Inductances L1 , L2 ,….and Ln connected in
parallel is given by :
1/ Leq = ( 1/ L1 + 1/ L2 + ….. 1/ Ln )
Note : The reciprocal of the equivalent inductance is the sum of the reciprocals of the
inductances.
Problem: Determine the equivalent inductance in the given network.
Solution: Calculating the first inductor branch L
A ,
where an Inductor L
5
in parallel with
inductors L
6
and L
7.
LA =
L5 × (L6 + L7)
L5 + L6 + L7
=
50mH × (40mH + 100mH)
50mH + 40mH + 100mH
= 36.8mH
Calculating the second inductor branch L
B
, where Inductor L
3
in parallel with
inductors L
4
and L
A
.
LB =
L3 × (L4 + LA)
L3 + L + LA
=
30mH × (20mH + 36.8mH)
30mH + 20mH + 36.8mH
= 19.6mH
Calculate the equivalent circuit inductance
LEQ,
where Inductor L
1
in parallel with
inductors L
2
and L
B
.
Malla Reddy College of Engineering and Technology (MRCET)
Department of EEE ( 2017-18 )
Electrical Circuits EEE LEQ, =
L1×(L2+LB)
L1+L2+LB
=
20𝑚𝐻×(40𝑚𝐻+19.6𝑚𝐻)
20𝑚𝐻+40𝑚𝐻+19.6𝑚𝐻
= 15𝑚𝐻
