Question

1. the total inductance in this circuit is

a 0.17 mh
b 1.2 mh
c 2.1 mh
d 3 mh

Explanation:

Step1: Analyze the connection of inductors

The inductors \(L_1 = 1\ \text{mH}\) and \(L_2=0.2\ \text{mH}\) are in parallel.

Step2: Apply the formula for parallel inductors

The formula for the equivalent inductance \(L_{eq}\) of two inductors \(L_1\) and \(L_2\) in parallel is \(\frac{1}{L_{eq}}=\frac{1}{L_1}+\frac{1}{L_2}\).
Substitute \(L_1 = 1\ \text{mH}\) and \(L_2 = 0.2\ \text{mH}\) into the formula:
\(\frac{1}{L_{eq}}=\frac{1}{1}+\frac{1}{0.2}=1 + 5=6\).
Then \(L_{eq}=\frac{1}{6}\approx0.17\ \text{mH}\).

Answer:

A. \(0.17\ \text{mH}\)