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A Review Part C IP Consider the circuit shown in the figure, which contains a 6.2.V battery, a 44-mH inductor, and four 60- r

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Answer #1

C) Req = 2*R*R/(2*R + R) + R

= 5*R/3

= 5*60/3

= 100 ohms

maximum current through the circuit,

I_max = V/Req

= 6.2/100

= 0.062 A

current through the inductor at t = T,

I = I_max*(1 -e^(-t/T))

= 0.062*(1 - e^(-T/T))

= 0.03919 A

Energy stored in the inductor, U = (1/2)*L*I^2

= (1/2)*0.044*0.03919^2

= 33.8*10^-6 J

= 33.8 micro J <<<<<<<<<<-------------Answer

D) after a longt time the switch is closed

Energy stored in the inductor, U = (1/2)*L*I_max^2

= (1/2)*0.044*0.062^2

= 84.6*10^-6 J

= 84.6 micro J <<<<<<<<<<-------------Answer

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