Question

Two inductors 0.4 H and 20.6 H are connected in parallel. If this combination is connected in series with an inductor of inductance 0.76 H. The equivalent inductance of the circuit will be (nearly)

• A. 2 H
• B. 1 H
• C. 0.2 H
• D. 0.1 H

Answer 1

The correct option is B 1 H

$L^1=\frac{L_1{L}_2}{L_1+L_2}+L_3$
$=\frac{0.4\times 20.6}{0.4+20.6}+0.76$
$=\frac{8.24}{21}+0.76$
$=0.39+0.76$
$=1.15$
$\approx 1H$