Question

1. the total inductance in this circuit is

a 3 mh
b 2 mh
c 1 mh
d 0.3 mh

Explanation:

Step1: Calculate total inductance for parallel inductors

For inductors in parallel, the formula is $\frac{1}{L_{total}}=\frac{1}{L_1}+\frac{1}{L_2}+\frac{1}{L_3}$. Given $L_1 = L_2 = L_3=6\space mH$.
So, $\frac{1}{L_{total}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$.

Step2: Simplify the equation

$\frac{1}{L_{total}}=\frac{1 + 1+ 1}{6}=\frac{3}{6}=\frac{1}{2}$.
Then, $L_{total}=2\space mH$.

Answer:

B. 2 mH