Question

1. the total inductance in this circuit is

a 169 h
b 44 h
c 1.65 h
d 0.044 h

Explanation:

Step1: Convert units

Convert \(L_3 = 1000\space mH\) to \(H\). Since \(1\space H= 1000\space mH\), then \(L_3 = 1\space H\).

Step2: Calculate inductance for parallel - connected \(L_2\) and \(L_3\)

For two inductors \(L_a\) and \(L_b\) in parallel, the equivalent inductance \(L_{eq1}=\frac{L_a\times L_b}{L_a + L_b}\). Here \(L_a = L_2=0.05\space H\) and \(L_b = L_3 = 1\space H\). So \(L_{eq1}=\frac{0.05\times1}{0.05 + 1}=\frac{0.05}{1.05}\approx0.0476\space H\).

Step3: Calculate total inductance (series - parallel)

Now \(L_1 = 0.6\space H\) is in parallel with \(L_{eq1}\). The formula for two inductors in parallel is \(L_{total}=\frac{L_1\times L_{eq1}}{L_1+L_{eq1}}\). Substitute \(L_1 = 0.6\space H\) and \(L_{eq1}\approx0.0476\space H\) into the formula: \(L_{total}=\frac{0.6\times0.0476}{0.6 + 0.0476}=\frac{0.02856}{0.6476}\approx0.044\space H\).

Answer:

d. \(0.044\space H\)