Question 284: Electromagnetic Induction

Question

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

Accepted Answer

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

Answer

0.0333

Answer

0.0667

Answer

0.0726

Answer

0.0984

Electromagnetic Induction

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

Choose the correct answer: