Question 195: Electromagnetic Induction

Question

Two inductors of 80 mH each are joined in parallel. The current passing through the combination is 2.1 A . The energy stored in this combination of inductors is

Accepted Answer

$$ L_1=L_2=L=80 \mathrm{mH} $$
When two inductors are connected in parallel, their equivalent inductance is given by,
$$ \frac{1}{L_{e q}}=\frac{1}{L_1}+\frac{1}{L_2} $$
$$ \begin{array}{ll} 	herefore & \mathrm{L}_{\mathrm{eq}}=\frac{\mathrm{L}}{2}=40 \mathrm{mH} \ & \mathrm{U}_{\mathrm{B}}=\frac{1}{2} \mathrm{~L}_{\mathrm{eq}} \mathrm{I}^2 \ 	herefore & \mathrm{U}_{\mathrm{B}}=\frac{1}{2} 	imes 40 	imes 10^{-3} 	imes 2.1 	imes 2.1 \ 	herefore & \mathrm{U}_{\mathrm{B}}=0.0882 \mathrm{~J}=8.82 	imes 10^{-2} \mathrm{~J} \end{array} $$

Answer

$4.84 	imes 10^{-2} \mathrm{~J}$

Answer

$7.26 	imes 10^{-2} \mathrm{~J}$

Answer

$8.82 	imes 10^{-2} \mathrm{~J}$

Answer

$10.85 	imes 10^{-2} \mathrm{~J}$

Electromagnetic Induction

Two inductors of 80 mH each are joined in parallel. The current passing through the combination is 2.1 A . The energy stored in this combination of inductors is

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