Assignment8_Solutions

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

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online