Assignment8_Solutions

Assignment8_Solutions - P43" The soluTion is of The...

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Unformatted text preview: P43" The soluTion is of The form of EquaTion 4.17: I pm) 2 i; + K2 exp(— and} 2 10 + 1:; exp{— rm x10_3)l in which K2 is a consTanT To be deTermined. AT T = 0 + , we have vfl0 +) = —10 =10 + K2 Solving, we find ThaT KB 2 —20 and The soluTion is V511"): 1o — 20 exp(— ma x10_3]) v SeTTing The voITage equal To zero aT Time To and solving we obTain: o = 10 — 20 exp{— r0112 x10'3)) 1/2 = expli— rO/(z x10'3)) —|n(2) = 40 /(2 X103] To : 2|n(2)=1.386 ms P4.4* The iniTiaI energy is w;=%c‘{i4)2 =%100x1o-6 x10002 = 50J AT T 2 3}, half of The energy remains, and we have 25 zéchfleHZ , which yields vh‘z ) = 707.1% The volTage across The capaciTance is given by v: {7‘} = K expl— #RC}: 1000 exp(— 103‘] for T > 0 SubsTiTuTing, we have 707.1 2 1000 expl— 101'2). Solving, we obTain |n(0.7071) = —101r‘2 r2 = 0.03466 seconds P4.11 (a) Rc-Zloms val?) = 10 for;T < 0 = 10 exp(— {10.01) = 10 expi— 1001‘) for T > 0 VR(T)=0 forT< 0 = 10 exp(— #001) = 10 expi— 1001‘) for f a 0 2 (b) p90") = [V931] = exp(— 2007‘) W for T > 0 p90") =0forT <0 P4.13 exp{— 200T)0’T — — (1 /200}exp(— 200?]; = 5 mJ (d) The iniTiaI energy stored in The capaciTance is w = gain-(0112 2%x100x10‘6x102 =5mJ Prior" To 3‘ = 0, we have 1250‘): 0 because The swiTch is closed. AfTer' f = 0, we can wriTe The following KCL equaTion aT The Top node of The circuiT: LIA?) ail/Cir} d? MulTipiying boTh sides by Rand subsTiTuTing values, we have 0/ r 0.02 ‘3: Hygirizzo (1) The soluTion is of The form um} = £1 + K2 exp[— r5196] = £1 + £2 expi— 50:9) (2) Using EquaTion (2) To subsTiTuTe inTo EquaTion (1), we evenTually obTain K1 = 20 The volTage across The capaciTance cannoT change insTanTaneously, so we have ‘ v€(0 ——}= vgio —]= 0 V€(0“}=0 =K/1 +K2 +6 =10mA P4.17 P4.21* Thus, K2 2 —K’1 = —20, and The soluTion is v1.41}: 20 — 20expl— 503‘] for' 3‘ > 0 The ske’rch should resemble The following plot: Ila-(f) (V) vensus f I . . : I I 0.03 0.05 0.06 0.07 0.08 0.09 I 0.02 o 0.61 0.04 0.1 The final vol’rage for each 2 s in’rerval is The initial volTage for The succeeding inTer'val. We have 3' = RC :1 s. For 0 g T g 2r we have v(r}=10 — 10 exp(—T}which yields v(2) 28.647 V. For- 2 3 T g 4, we have v(f) = 8.647 exp[—(3‘— 2}]which yields v(4) = 1.170 ‘v‘. For- 4 S f i 6, we have v0) =10 — (10 —1.170)exp[—(T — 4}]which yields V(6) = 8.805 V. Finally, for 6 3 T g 8, we have V0") 2 7.176 exp[—(f — 6)] which yields V(8)=1.192 V. In sTeady sta’re, The equivalent circuiT is: 900.11.. Thus, we have i; = O g=§=2A P424 Prior to f = 0, the steady-state equivalent circuit is: UV and we see that VG =12 V. A long time after 1' = 0, the steady-state equivalent circuit is: 12V 4O =8‘v‘. and we have v8 = 12 O + 20 P4.33* In steady state with the switch closed, we have iii] 2 O for T < 0 because the closed switch shorts the source. In steady state with the switch open, the inductance acts as a short circuit and the current becomes {inc} = 1A. The current is of the form fit] = K1 + K2 expl— Rfiélfor 1‘ 2 O in which 2? = 20 El, because that is the Thévenin resistance seen looking back from the terminals of the inductance with the switch open. Also, we have ‘ r'(0+_]= r'lO—Jz 0 2 K1 +K2 rim) 2 1 2 K1 Thus, K2 2 —1 and the current (in amperes) is given by {(1‘) = 0 for f c O 21— exp(— 20?) for t 2 O P439 In sTeady sTaTe, The inducTor acTs as a shor‘T circuiT. WiTh The swiTch closed, The sTeady—sTaTe currenT is (20 Nil/"(5 Q) = 4 A. WiTh The swiTch openedr The currenT evenTually approaches {Joe} 2 (20 Rib-T20 Q) = 1A. For' f > O, The currenT has The form Mr} = K1 + K2 6XP(- 97/1) where R = 20 9, because ThaT is The r'esisTance wiTh The swiTch open. Now, we have ;;(0+]=x;(o—)=4=Kl+K2 ;;(m)=1=K1 Thus, we have K2 2 3. The currenT is {LU} = 4 f < O (swiTch closed) 21+ 3exp(—10T) T 2 O (swiTch open) IL versus f I ‘ I —0.1 o 0.1 0.2 as 0.4 0.5 ...
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Assignment8_Solutions - P43" The soluTion is of The...

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