918_Physics ProblemsTechnical Physics

918_Physics ProblemsTechnical Physics - 258 P32.69...

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258 Inductance P32.69 Left-hand loop: ε −+ = IIR IR 21 2 2 0 bg . Outside loop: = IIR L dI dt 21 0 . Eliminating I 2 gives ′− = IR L dI dt 0. This is of the same form as Equation 32.6, so its solution is of the same form as Equation 32.7: FIG. P32.69 It R e RtL af ej = −′ 1 . But ′ = + R RR 12 and ′ = + R 2 , so = + + = R RRR R R R 1 2 1 . Thus R e a f =− 1 1. P32.70 When switch is closed, steady current I 0 = 1.20 A . When the switch is opened after being closed a long time, the current in the right loop is II e = 0 2 so e I I Rt L = 0 and Rt L I I = F H G I K J ln 0 . Therefore, L Rt == = = 2 0 100 0150 120 00956 956 ln .. ln . a f s A 0.250 A H m H . FIG. P32.70 P32.71 (a) While steady-state conditions exist, a 9.00 mA flows clockwise around the right loop of the circuit. Immediately after the switch is opened, a 9.00 mA current will flow around the outer loop of the circuit. Applying Kirchhoff’s loop rule to this loop gives:
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