Question

1. the total inductance in this circuit is

a 1650 h
b 165 h
c 16.5 h
d 1.65 h

Explanation:

Step1: Convert units

Convert \(L_1 = 600\space mH\) to \(H\): \(L_1=600\times10^{- 3}\space H = 0.6\space H\)
Convert \(L_3 = 1000\space mH\) to \(H\): \(L_3 = 1000\times10^{-3}\space H=1\space H\)

Step2: Calculate total inductance

For inductors in series, \(L_{total}=L_1 + L_2+L_3\)
Substitute \(L_1 = 0.6\space H\), \(L_2 = 0.05\space H\), \(L_3 = 1\space H\)
\(L_{total}=0.6 + 0.05+1=\space 1.65\space H\)

Answer:

d. \(1.65\space H\)