Two inductors of 60 mH each are joined in par | vedclass

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(C) Given: $L_1 = L_2 = L = 60 mH = 60 	imes 10^{-3} H$.Current $I = 2.2 A$.When two inductors are connected in parallel,the equivalent inductance $L

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$0.0726$

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(C) Given: $L_1 = L_2 = L = 60 mH = 60 	imes 10^{-3} H$.Current $I = 2.2 A$.When two inductors are connected in parallel,the equivalent inductance $L_{eq}$ is given by:$\frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2} = \frac{1}{L} + \frac{1}{L} = \frac{2}{L}$.Therefore,$L_{eq} = \frac{L}{2} = \frac{60 mH}{2} = 30 mH = 30 	imes 10^{-3} H$.The energy stored in an inductor is given by $U = \frac{1}{2} L_{eq} I^2$.Substituting the values:$U = \frac{1}{2} 	imes (30 	imes 10^{-3}) 	imes (2.2)^2$.$U = 0.5 	imes 30 	imes 10^{-3} 	imes 4.84$.$U = 15 	imes 10^{-3} 	imes 4.84 = 0.0726 J$.

Two inductors of $60 mH$ each are joined in parallel. The current passing through this combination is $2.2 A$. The energy stored in this combination of inductors in joule is

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