The inductors of two LR circuits are placed n | vedclass

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(C) In the steady state,the currents in the two circuits are:$I_1 = \frac{V_1}{R_1} = \frac{5 	ext{ V}}{5 	ext{ }\Omega} = 1 	ext{ A}$$I_2 = \frac{

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$55$

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(C) In the steady state,the currents in the two circuits are:$I_1 = \frac{V_1}{R_1} = \frac{5 	ext{ V}}{5 	ext{ }\Omega} = 1 	ext{ A}$$I_2 = \frac{V_2}{R_2} = \frac{20 	ext{ V}}{10 	ext{ }\Omega} = 2 	ext{ A}$The total energy stored in the system of two coupled inductors is given by:$U = \frac{1}{2} L_1 I_1^2 + \frac{1}{2} L_2 I_2^2 + M I_1 I_2$Substituting the given values $(L_1 = 10 	ext{ mH}, L_2 = 20 	ext{ mH}, M = 5 	ext{ mH}, I_1 = 1 	ext{ A}, I_2 = 2 	ext{ A})$:$U = \frac{1}{2} 	imes (10 	imes 10^{-3}) 	imes (1)^2 + \frac{1}{2} 	imes (20 	imes 10^{-3}) 	imes (2)^2 + (5 	imes 10^{-3}) 	imes 1 	imes 2$$U = 5 	imes 10^{-3} + 40 	imes 10^{-3} + 10 	imes 10^{-3}$$U = 55 	imes 10^{-3} 	ext{ J} = 55 	ext{ mJ}$

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