Two inductors of $60 \mathrm{mH}$ each are joined in parallel. The current passing through this combination…
Two inductors of $60 \mathrm{mH}$ each are joined in parallel. The current passing through this combination is $2.2 \mathrm{~A}$. The energy stored in this combination of inductors in joule is
$0.0333$
$0.0667$
$0.0726$
$0.0984$
Solution
$\mathrm{L}_1=\mathrm{L}_2=\mathrm{L}=60 \mathrm{mH}$
When two inductors are connected in parallel, their equivalent inductance is given by,
$\begin{aligned}
& \frac{1}{\mathrm{~L}_{\text {eq }}}=\frac{1}{\mathrm{~L}_1}+\frac{1}{\mathrm{~L}_2} \\
& \therefore \quad \mathrm{L}_{\mathrm{eq}}=\frac{\mathrm{L}}{2}=30 \mathrm{mH} \\
& \mathrm{u}_{\mathrm{B}}=\frac{1}{2} \mathrm{~L}_{\mathrm{eq}} \mathrm{I}^2 \\
& \therefore \quad \mathrm{u}_{\mathrm{B}}=\frac{1}{2} \times 30 \times 10^{-3} \times 2.2 \times 2.2 \\
& \therefore \quad \mathrm{u}_{\mathrm{B}}=0.0726 \mathrm{~J} \\
&
\end{aligned}$