Basic Electrical and Electronics Engineering: Unit I: Electrical Circuits
Grouping of Inductors
with Solved Example Problems | Electrical Circuits
Two inductances L1 and L2 are connected in series with a voltage V.Here the supply voltage is equal to the summation of voltage drops across L1 and L2.
GROUPING
OF INDUCTORS
Two
inductances L1 and L2 are connected in series with a
voltage V as shown in fig 1.18
Here the supply voltage is equal to the summation of voltage drops across L1 and L2.
L
= L1 + L2
L
is the equivalent inductance.
If
two inductances L1 and L2 are connected in parallel, the
supply voltage is V1 the branch currents are i1 and i2
as shown in fig 1.19.
The
supply voltage is equal to the voltage drop across each and every inductance
From
fig 1.19
i
= i1 + i2
Example: 12
The inductances of 4H each are
connected in series. What is the equivalant
inductance?
Equivalent
inductance ⇒
L = L1 + L2 + L3
L
= 4+4+4
=
12Ω
Example: 13
The inductances are connected as
shown in fig. Find the equivalent inductance.
4H
and 4H are connected in parallel
The
net inductance across A-B is
LA-B
= L1 + L2 + L3
=
2 + 2 + 1
=
5H
Basic Electrical and Electronics Engineering: Unit I: Electrical Circuits : Tag: : with Solved Example Problems | Electrical Circuits - Grouping of Inductors
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