Question

Pure inductors each of inductance $3\text{H}$ are connected as shown. The equivalent inductance of the circuit is

• A. $1\text{H}$
• B. $2\text{H}$
• C. $3\text{H}$
• D. $9\text{H}$

Answer 1

The correct option is A $1\text{H}$

The given circuit can be redrawn as follows

All the three inductors are in parallel then their equivalent is given by

$\frac{1}{L_{eq}}=\frac{1}{L_1}+\frac{1}{L_2}+\frac{1}{L_3}+....$

$\frac{1}{L_{eq}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}$

$L_{eq}=1\text{H}$