PHYS
9 solution

# 9 solution - DiS'eussi0n.QuestioI1 10A P2_1.21.We'e1 10 M C...

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Unformatted text preview: DiS'eussi0n.QuestioI1 10A P2'_1.21,_.We'e1{ 10. M. C 11011113 The d1agram shOWs a classic “LR” circuit cor_1t._a_ining_Th both . E: 10 V resistors and inductors lg. ' L_ — 15 mH sW1tch shoWI11s 1111t1allyTh R1 = 4 Q connected to 11631115161 terminal. _ . R2 : 6- Q and 1s then throWn to position “ a” at time. t— 0. (a) "At-t = 0+1. just .-after the switch is 'thrmn to position a, what-are the currents I1 and Ii:- across the. two resistors? Just after-theswitch-‘is thro-Wn, what does the inductor *look like” to. the res-gt of the circuit? What is the current doing? Do inductors like'that? Once ”you know What the inductor looks like at this '-po.int...inrtim.e,- .it. is highly advisable to redraW the circuit. Just before-the SWit'ch is throWn, there is no current in the inductor, since the inductor blocks immediate changes in current, the current Will he still be zero; Hence the-same current Will ﬂoW through R1 and. R2. The voltage drop through both resistors Will equal 5 . Hence 5 10 _ =—=1A :R1+R2 4+6 (b) After a. Very long time, what is th'e'i'ns'tantaneons p'..0W-e-r-"dissip rated in; the. circuit? After a Very" long time-1 What will have happened to 1116- eurrent? NO-W'What will the inductor look. like to the rest of the circuit? The circuit n20W takes- on a simple form What current is ﬂowing through the resistors? After a long time, the rate of change of current has .-'stahi'lizejd and hence AV: -Ld—I— - O. This means no current ﬂoWs through R. d1 10 - , - _' .—=§- . Power =1: R2. = 2 6 3 3 g 2 Hence 6 = 12R; —> JEFF: ] 6:16.66 W (1:). Sketch the behawor of the energy stored' 111 the; Inductor as a function of tinjle. What 15 the ﬁnal energy stored after a very long time? Think about the current through. the inductor: What is it attime 0.? after 'agv-ery long time?- Initially the inductor blocks the current through it. The current then ﬂows according to If 2 Which resembles the 1(1) curate. After a longtime I=lﬂm (l-exp(—t/1')) The stored energy-is U = 1 - 5- 2 =_(1-_5 mH)[— A] = 20.83111J .2' 3 ...
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