Two inductors of 80 mH each are joined in pa | vedclass

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(C) Given: Inductance of each inductor $L_1 = L_2 = 80 \ mH = 80 	imes 10^{-3} \ H$.Since the inductors are connected in parallel,the equivalent indu

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

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(C) Given: Inductance of each inductor $L_1 = L_2 = 80 \ mH = 80 	imes 10^{-3} \ H$.Since the inductors are connected in parallel,the equivalent inductance $L_{eq}$ is given by $\frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2}$.$L_{eq} = \frac{L_1 	imes L_2}{L_1 + L_2} = \frac{80 	imes 80}{80 + 80} \ mH = \frac{6400}{160} \ mH = 40 \ mH = 40 	imes 10^{-3} \ H$.The current passing through the combination is $I = 2.1 \ A$.The energy stored in an inductor is given by $U = \frac{1}{2} L_{eq} I^2$.$U = \frac{1}{2} 	imes (40 	imes 10^{-3}) 	imes (2.1)^2$.$U = 20 	imes 10^{-3} 	imes 4.41$.$U = 88.2 	imes 10^{-3} \ J = 8.82 	imes 10^{-2} \ J$.

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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