HW3 Solution

# HW3 Solution - P4.9 Prior‘ To f O we have v(T 0 because...

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Unformatted text preview: P4.9 Prior‘ To f : O, we have v(T) : 0 because The swiTch is closed. AfTer‘ T : 0, we can wr'iTe The following KCL equaTion (IT The Top node of The cir‘cuiT: @wdvlﬂzmxi 2 df MulTiplying boTh sides by Rand subsTiTuTing values, we have cit/(T) 0.01 + T : 10 1 df v{ ) ( ) The soluTion is of The form Mr) : K1 + K2 exp(— f/RC'): Kl + K2 exp(—100r) (2) SubsTiTuTing EquaTion (2) inTo EquaTion (1) we evenTually obTain K1 :10 The volTage across The capaciTance cannoT change insTanTaneously, so we have 1/(0 + :v(0—):O Thus, Kg 2 —/(1 = —10, and The soluTion is v{T):10 e 10 exp(7100f)for' f > O P4.16 The final volTage for each ls inTer‘val is The iniTial volTage for' The succeeding inTer‘val. We have 7 : 96' :1 s. For‘ 0 s f 31, we have vU) : 10 i 10 expGT) which yields v(1) : 6.321 V. For' 1: f S 2, we have v(2‘) : 6.321e>~:p[—(aL —1)]which yields v(2) : 2.325 V, For‘ 2 g f g 3, we have v(7‘) : 10 —(10 — 2.325) expi—(r —2)]which yields v(3) : 7.176 v. Finally, for 3 s T s: 4, we have v0“) : 7.176 exp[i(f * 3)]which yields l/(4) = 2.640 v. P4.17 (a) The volTages across The capaciTor‘s cannoT change insTanTaneously. Thus, v1(0 +) : v1(O —) : 100 V and V2(0 +) : v2{0 —) : 0. Then, we can wr‘iTe {(0 i)=V1l0+l—V2lo+l= 100—0 :lmA R 100x103 ('0) Applying KVL, we have — V1(?‘)+ REG) + 1/2 (3‘) = O 1 f. . 1 f. — I(f)df—100+Rl(f)+— 1(f)a’f+O:O cal cal Taking a derivaTive wiTh respecT To Time and rearranging, we obTaI'n . l/ \ o’;(fl+l|i+ijg(r):o (1) dr 9 Kc; 6; . 5152 :50 . CHLCE ms (c) The Time consTanT is r = R (d) The soluTion To EquaTion (1) is of The form M) = K1 expl- Hr) However, ITO +):1mA, so we have K1 :1mA and {(f) : exp(— 20?“) mA. (e) The final value of v20) is v2(m) z Cilandr + v, (0 +) 20 , =106j1o-3 expl— r/0.05)dr + o 0 2103(— 0.05)exp(— 270.0513 : 50 V Thus, The iniTial charge on 6; is eventually divided equally beTween Clandé'z. P422 In sTeady sTaTe wiTh a dc source, The inducTance acTs as a shorT circuiT and The capaciTance ads as an open circuiT. The equivalenT circuiT is: P436 L1 = (100 V)/(1kQ) : 100 mA {3 : O {a : (100 V)/(1kQ) : 100 mA {i={é+lé+!;1=200mA V5 : 100 V In sTeady sTaTe, The inducTor‘ ads as a shor'T cir‘cuiT. WiTh The swiTch open, The sTeady—sTaTe cur'r'enT is (100 V)/(100 Q):1A. WiTh The swiTch closed, The cur‘r‘enT evenTually approaches = (100 V)/(25 Q) = 4 A. For' f > O, The cur‘r‘enT has The form {(r) : K1 + K2 exp(— RT/i’.) where R : 25 Q, because ThaT is The r‘esisTance wiTh The swiTch ciosed. Now, we have J'(O+)= r‘(O—)= 1 : K1 +K2 {(00) : 4 : Kl Thus, we have K2 : —3. The cur'r'enT is {(1‘) :1 T <; O (swiTch open) : 4 — 3 expi—125f] f O (swiTch closed) ﬁt) an *---—z-- - — —- 1* (Ins) 30 I90 62% P438 Prior‘ To 7* : 0, The cur‘r‘enT source is shor‘Ted, so we have 1'10“) : O for 7‘ <4 0 AfTer' The swiTch opens oT 3‘ : O, The cur'r'enT {10‘} increases from zero headed for 1 A, The inducTonce sees a Thévenin resisTonce of 4 Q; and The Time consTonT is r : L/4 :1 s, so we have {1(7‘) :1— exp(—T) for O a 7‘ <1 AT T :1, The curr‘enT r‘eoches 0632 A. Then, The swiTch closes, The source is shor‘Ted, and The currenT decoys Toward zero. Because The inducTonce sees a r‘esisTonce of 2 Q, The Time consTonT is r = L/Z : 2 3. {1(7‘) : O.632exp(—T/2) for1< T E4.3 (a) In dc sTeody sToTe, The copociTonces ocT as open cir‘cuiTs and The inducTonces ocT as shor‘T circuiTs. Thus The sTeody-sToTe (i.e., r approaching infiniTy) equivolenT cir'cuiT is: 4. 99v; _. From This cincuiT, we see ThoT r; : 2 A. Then ohm‘s law gives The volToge as V“ : Ric : 50 V‘ (b) The dc sTeady-sToTe equivalenT cincuiT is: Here The Two 10—!) r'esisTances one in parallel wiTh on equivolenT r‘esisTonce of 1/(1/10 + 1/10) = 5 Q. This equivalenT r‘esisTonce is in series wiTh The 5-1] resisTonce. Thus The equivolenT resisTonce seen by The source is 10 Q, and ti : 20/10 : 2 A. Using The cur‘r‘enT division principle, This cur‘r‘enT spliTs equally beTween The Two lO-ﬂ nesisTonces, so we have {2 : f3 : 1 A. E4.4 (a) 7: 1/22 : 0.1/100 :lms ('0) Just before the swiTch opens: the circuit is in dc steady state wiTh an inductor current of K /R1 = 1.5 A. This current continues to flow in The inductor immediately after the switch opens so we have f(O+) : 1.5 A. This current must flow (upward) through R2 so the initial value of The voltage is v(0+) : —R2r'(0+) : —150 V. (c) We see that The iniTial magnitude of (f) is ten Times larger Than The source voltage. (d) The voltage is given by Va) : — V51 exp(—r / r) : —15o exp(—1000t) RIF LeT us denote The Time at which the voltage reaches haltc of its initial magnitude as to. Then we have 0.5 : exp(—10007‘H) Solving and substituting values we obtain fH : —1O‘3ln(0.5) :10‘3ln(2): 0.6931 ms ...
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HW3 Solution - P4.9 Prior‘ To f O we have v(T 0 because...

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