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MHT CET · Physics · Alternating Current

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

  1. A \(0.0333\)
  2. B \(0.0667\)
  3. C \(0.0726\)
  4. D \(0.0984\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(0.0726\)

Step-by-step Solution

Detailed explanation

\(
\mathrm{L}_1=\mathrm{L}_2=\mathrm{L}=60 \mathrm{mH}
\)
When two inductors are connected in parallel, their equivalent inductance is given by,
\(
\begin{aligned}
& \frac{1}{\mathrm{~L}_{\text {eq }}}=\frac{1}{\mathrm{~L}_1}+\frac{1}{\mathrm{~L}_2} \\
& \therefore \quad \mathrm{L}_{\mathrm{eq}}=\frac{\mathrm{L}}{2}=30 \mathrm{mH} \\
& \mathrm{u}_{\mathrm{B}}=\frac{1}{2} \mathrm{~L}_{\mathrm{eq}} \mathrm{I}^2 \\
& \therefore \quad \mathrm{u}_{\mathrm{B}}=\frac{1}{2} \times 30 \times 10^{-3} \times 2.2 \times 2.2 \\
& \therefore \quad \mathrm{u}_{\mathrm{B}}=0.0726 \mathrm{~J} \\
&
\end{aligned}
\)
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