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The pure inductors each of inductance $6 \mathrm{H}$ are connected as shown in the figure. Their equivalent inductance between the points ' $\mathrm{P}$ ' and ' $\mathrm{Q}$ ' is
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The correct answer is:
$2 \mathrm{H}$
In the given circuit, one end of each inductor is joined together. Similarly, the other end of each inductor is joined together. Hence the three inductors are connected in parallel. Their equivalent inductance is given by
$$
\begin{aligned}
& \frac{1}{L}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6} \\
& \therefore \mathrm{L}=2 \mathrm{H}
\end{aligned}
$$
$$
\begin{aligned}
& \frac{1}{L}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6} \\
& \therefore \mathrm{L}=2 \mathrm{H}
\end{aligned}
$$
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