The inductor arrangement of Figure, with

L1=30.0mH,L2=50.0mH,L3=20.0mHandL4=15.0mHis to be connected to a varying current source. What is the equivalent inductance of the arrangement?

Short Answer

Expert verified

Leq=59.3mH

Step by step solution

01

Given

L1=30.0mH;L2=50.0mH;L3=20.0mH;L4=15.0mH

02

Understanding the concept

Equivalent inductance forinductors connected in series isthesum of all inductances. And reciprocal of equivalent inductance for parallel connection of inductances isthesum of reciprocals of inductances.

Formulae:

Equivalent inductance forinductors connected inparallel is

1Leq=n=1N1Ln

Equivalent inductance forinductors connected inseries is

Leq=n=1NLn

03

Calculate the equivalent inductance of the arrangement

L2and L3are connected in parallel, so their equivalent is given by

1L23=1L2+1L31L23=150+120

Hence,

L23=14.3mH

Now, L23,L1andL4are in series.

Leq=L1+L23+L4Leq=30+14.3+15Leq=59.3mHz

Therefore the equivalent inductance = 59.6mH

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