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<Assignment10 Alternative Exercise 23.107 7 of 25 Review IP Consider the circut shown in the figure, which contains a 70-V ba
<Assignment10 Alternative Exercise 23.107 IP Consider the circuit shown in the figure, which contains a 7.0-V battery, a 45-m
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Answer #1

The total resistance is

r=R+(1/R+1/2R)^{-1}=5R/3 = 5 \times 57/3 =95\,\Omega

the current in the circuit 1 characteristic time interval after the switch is closed is

I=V/r(1-e^{-1})=7/95(1-0.368)=0.0466\, A

The energy stored in the circuit 1 characteristic time interval after the switch is closed is

U=LI^2/2=0.045\times 0.0466^2/2=48.81\times 10^{-6}J = 48.81\,\mu J

The current in the circuit in steady state is

I=V/r=7/95=0.0737\,A

The energy stored in the inductor after a long time is

U=LI^2/2=0.045\times 0.0737^2/2==122.2\times 10^{-6}J = 122.2\,\mu J

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