ProbSolv_Chapter07

ProbSolv_Chapter07 - CHAPTER 7 - FIRST-ORDER CIRCUITS List...

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Unformatted text preview: CHAPTER 7 - FIRST-ORDER CIRCUITS List of topics for this chapter : Source-Free RC Circuit Source—Free RL Circuit Singularity Functions Step Response of an RC Circuit Step Response of an RL Circuit First-Order Op Amp Circuits Transient Analysis with PSpice Applications SOURCE-FREE RC CIRCUIT M Problem 7.1 For the circuit in Figure 7 .1, find VC (t) and iC (t) given VC (0) = 10 V . 10 Q . ic(t) + , vat) “5 F Figure 7.1 > Carefully DEFINE the problem. Each component is labeled completely. The problem is clear. > PRESENT everything you know about the problem. This is a source-free RC circuit. The natural response of this source-free RC circuit is vC (t) = VO 6"“ , where V0 = VC (0) and t = RC We know the initial voltage across the capacitor. To find the capacitor voltage for any time greater than zero, we need to calculate the time constant of the circuit. > Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success. The three solution techniques that can be used are nodal analysis, mesh analysis, and basic circuit analysis. Basic circuit analysis can be used to solve this problem. 109 > ATTEMPT a problem solution. V0 =vc(0)=10V and r =RC=(10)(1/5)=23 VC(t)=106‘t/2 V . - V (t) . dv t 1C0) = I: or 1C(t) = C (is ) In either case, iC (t) = -e't/2 A > EVALUATE the solution and check for accuracy. Using KVL, 10ic(t) +\vC(t) = (10)(_e-t/2) + 1064/2 ___ 0 Our check for accuracy was successful. > Has the problem been solved SATISFACTORILY? If so, present the solution; if not, then return to “ALTERNATIVE solutions” and continue through the process again. This problem has been solved satisfactorily. i C(t) = —e“'/2 amps for all t >0 . Problem 7.2_ [7.23] Express the signals in Figure 7.2 in terms of singmlarity functions. VICE) V2(t) Figure 7.2 110 (a) v1 (t) = u(t +1) — u(t) + [u(t — 1) — u(t)] V1 (t) = u(t + 1) — 2 u(t) + u(t —— l) (b) v20) = (4 —t>[u<t —2) —u(t — 4)] v2(t) = -(t—4)u(t—2)+(t—4)u(t—4) V2(t) = 2u(t—2)—r(t—2)+r(t—4) (c) V3(t)=2[u(t—2)-u(t-4)]+4[u(t—4)—u(t—6)] v3(t) = 2u(t—2)+2u(t—4)—4u(t—6) (d) v4(t)=-t[u(t—1)—u(t—2)]=-tu(t—1)+tu(t—2) V4(t)=(-t+l—l)u(t—1)+(t—2+2)u(t—2) v4(t)=-r(t—l)—u(t—l)+r(t—2)+2u(t—2) Problem 7.3 Given i(t) = 3 6'“2 A , find VC (t) for the circuit shown. in Figure 7 .3. 10 Q 1/5 F Figure 7.3 vC (t) = 30e-t/2 V Problem 7.4 Given VC (1) = 10 V , find VC (t) for all t > 0 in Figure 7.4. 109. ' + veg) 1/20 F Figure 7.4 10 at Vc(t)=[-ej)e V 111 OW SOURCE—FREE RL CIRCUIT Problem 7.5 For the circuit in Figure 7.5, find i(t) and vL (t) given i(0) = 4 A. 10 Q i“) + vL(t) E 5 H Figure 7.5 > Carefully DEFINE the problem. Each component is labeled completely. The problem is clear. > PRESENT everything you know about the problem. This is a source-free RL circuit. The natural response of this source—free RL circuit is iL(t) = 10 6"“, where 10 = iL(0) and z = L/R We know the initial current through the inductor. To find the current through the inductor for any value of time greater than zero, we need to calculate the time constant of the circuit. > Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success. The three solution techniques that can be used are nodal analysis, mesh analysis, and basic circuit analysis. Basic circuit analysis can be used to solve this problem. ' > ATTEMPT a problem solution. IO=iL(0)=4A and t=L/R=5/10=O.5s i(t)= 4e'2‘ A d. VL(t) = ~10i(t) or VL(t) = L? In either case, vLa) = -40e'2‘ v 112 > EVALUATE the solution and check for. accuracy. . Using KVL, 10i(t) + vL (t) : (1o)(4e-2t) ~ 406—2: = 0 Our check for accuracy was successful. > Has the problem been solved SATISFACTORILY? If so, present the solution; if not, then return to “ALTERNATIVE solutions” and continue through the process again. This problem has been solved satisfactorily. i(t) = 4e'2‘u(t) A and VL (t) = -40 e'“u(t) V Problem 7.6 For the circuit in Figure 7.6, find i(t) given VL (t) = 20 e'2t V . 20 Q i“) + vL(t) E 10H Figure 7.6 L/R=10/20= 0.5 5 KO = % Ivar) dc but it is also - v t i(t) = L( ) = -e‘2'u(t) A R __.____ Problem 7.7 Given i(0) = 2 A , find i(t) , p109 (power absorbed by the 10 ohm resistor), and w109 (total energy dissipated by the 10 ohm resistor) for the circuit in Figure 7.7. 10 Q i“) 5 H Figure 7.7 113 i(t) = 2e'2t A pIOQ = 40c”4t W wIOQ : 10 J SINGULARITY FUNCTIONS Problem 7.8 Solve for du(t) dr(t) (3) dt (b) dt 0 t<0 (a) HG):1 t>0 d j 0 t < 0 d Eua) = undefined t = 0 ———-> an“) 2 5(t) 0 t > 0 {0 t s 0 (b) r“) — t t > 0 d t {0 t s 0 d dtrU— 1 t>0 dtrm—fl Problem 7.9 Given VC (t) = [5 u(t) + 6r(t)] V , find ic(t) for the circuit in Figure 7.8. + Tic“) Figure 7.8 d icm = (:5;va 114 . ‘ 1 d l d . ic(t)=1—6'3;[5u(t)+6r(t)]=16.[52i’;u(t)+62§1;r(t):lI Using Problem 7.8, it is clear that 1 ice): 15[56(t)+ 6am] A Problem 7.10 Solve for (a) I8(t)dt (b) {new <20 359. 0» fl «2W STEP RESPONSE OF AN RC CIRCUIT Problem 7.11 Given V(t) = 20 u(t) V , find VC (t) and ic(t) in Figure 7.9. 100 1/20 F Figure 7.9 'c =RC=(10)(§16)=';-S vc(0)=ov vc(oo)=20V vC (t) = (20)(1 — e-2t )u(t) V d t 1 10a) = (3 V5: ) = (56)(-20)(—2e'2‘) = 2e'2‘ u(t) A 115 Problem 7.12 [7.37] Find the step responses V(t) and i(t) to vS = 5u(t) V in the circuit of Figure 7 . 10 129 79 Figure 7.10 Fort<0, Vs=5u(t)=0 —> V(O)=0V 0 5 V ( ) (5) 5 V = 00 : : - For t> , vS , v 4+12 4 Req=7+4|l12=10£2, T=ReqC=(10)(1/2)=SS «0 = V(oo) + [ v(0> — V(oo)] e-W V(t) = 1.25(1 — e"/5) V ' _ 9X _ _1~ 4/5 1(0‘ dt {2] 4 5 e i(t) = 0.125 e"/5 A Problem 7.13 Given V(t) = 10[u(t) —— u(t — 2)] V , find VC (t) in Figure 7.11. 109 1/10 F Figure 7.11 116 Find the Thevenin equivalent of the circuit at the terminals of the capacitor. This will simplify . the circuit, forming an RC circuit with a voltage source. Use the following circuit to find the open circuit voltage. 10 Q V0c must be equal to V(t) , since i(t) + 2i(t) = 0 ——> i(t) = 0 A‘ To find the short circuit current, 109 we a 0 15¢ . 1(1). 150 = i(t) + 21(1) = 31(1) where i(t = 1 0 I 3 (t) = —- v SC 1 0 Thus, R - Voc -— 19 Q “1 “ 156 ‘ 3 which leads to the following Thevenin equivalent circuit. 10/3 Q V(t) 1/ 10 F Using the Thevenin equivalent circuit with the capacitor as the load, we can see that 101 1 . TZR‘hC‘isi 10)”? 117 Fort<0, V(t)=0V, vC(0)=0V, Vc(oo)=0V vc(t)=OV For0<t<2, V(t)=10V, VC(O)=OV, V(oo)=10V Vc(t)=(10)(1—e‘3‘)V For2<t, v(t)=0V, vC(2)=(10)(1—e~6)v, vc(oo)=OV vC (t) = (10)(1— e'6)e'3("2) V Combining these cases, vC (t) = { [(10)(1 — e‘3' )[u(t) — u(t — 2)]]+ [9.975 e-“W u(t -— 2) ]} v STEP RESPONSE OF AN RL CIRCUIT Problem 7.14 Given V(t) = 40 u(t) V , find iL (t) and VL (t) in Figure 7.12. 10 Q V(t) 0 Figure 7.12 5H NH iL(0)=OA iL (t) = (4)(1 — e'“) “(t) A 'diL(t) dt vL (t) = L = (5)(4)(2)e'2‘ u(t) = 40 e-2t u(t) v 118 . Problem 7.15 [7. 55] Find V0 (t) for t > O in the circuit ofFigure 7.13. 6 Q 4H NH Figure 7.13 Let i be the inductor current. Fort < 0, the inductor acts like a short circuit and the 2 Q resistor is short-circuited so that the equivalent circuit is shown in Fig. (a). 69 _ i 69 i0 v i . 109 10o (a) > (b) i=i(0)=1—69=1.6667A L 4 Fort>0, Rm=2+3H6=4Q, T=EE=Z=1S To find i(oo) , consider the circuit in Fig. (b). 10—V v X v_i_0_V 2 a “6 i0) = i(oo) + [ i(m — i(oo)] e-t/T . 5 10 5 _t 5 4 t=—+ ———-—e =-—1+ A 10 6 [6 6) 6( e) 119 V0 is the voltage across the 4 H inductor and the 2 Q resistor di 10 10 _t (5) 4 10 10 _t Ldt—6+6e +(4)6(-1)e —6—6e vo(t)=1.6667(1—e")V vo(t):2i+ Problem 7.16 Find iL (t) and VL (t) in Figure 7.14 for V(t) = [20 u(t) -— 40 u(t ~ 1)] V . 10S) Figure 7.14 iL (t) = (2)(1 — e“)[u(t) — u(t —- 1)] + [-2 + (4 — 2e'l)e'““1)]u(t— 1) A vL (t) = 20c" 11(t)—[20e't + (20)(2 ~ e'1)e'(‘"”]u(t — 1) V _ . FIRST-ORDER OP AMP CIRCUITS Problem 7.17 Given V(t) = 10u(t) V , find i0 (t) for the circuit in Figure 7.15. 1 kg 1 mF 10 Q Figure 7.15 va — v(t) d +1ma‘;[va — Vo (t)] = 0 , 1k where Va=Vb=0V. 120 _vo(t) = -_[v(t) dt = -[10u(t)dt = -10t V io(t) = Viét) = —tu(t) A Problem 7.18 [7.59] Obtain V0 for t > 0 in the circuit of Figure 7.16. t = 0 Figure 7.16 This is a very interesting problem and has both an important ideal solution as well as an important practical solution. Let us look at the ideal solution first. Just before the switch closes, the value of the voltage across the capacitor is zero, which means that the voltage at both input terminals of the op amp are zero. As soon as the switch closes, the output tries to go to a voltage such that the inputs to the op amp both go to 4 volts. The ideal op amp puts out whatever current is necessary to reach this condition. An infinite (impulse) current is necessary if the voltage across the capacitor is to go to 8 volts in zero time (8 volts across the capacitor will result in 4 volts appearing at the negative terminal of the op amp). So V() will be equal to 8 volts for all t > 0. What happens in a real circuit? Essentially, the output of the amplifier portion of the op amp goes to whatever its maximum value can be. Then, this maximum voltage appears across the output resistance of the op amp and the capacitor that is in series with it. This then results in an exponential rise in the capacitor voltage to the steady-state value of 8 volts. I For all values of vC (t) less than 8 V, vC (t) = V op-amp-max (1 _. e't/(Routc)) V where Vop_m)_max is the maximum value of the op amp and Rout is the real output resistance of the practical op amp. When t is large enough so that the 8 V is reached, VC (t) = 8 V 121 TRANSIENT ANALYSIS WITH PSPICE Problem 7.19 The switch in Figure 7.17 moves from position a to b at t = 0. may Use PSpice to find i(t) for t > 0. a 6 Q 4 Q i“) 108 V O E 2 H Figure 7.17 (a) (b) When the switch is in position a, the schematic is shown below. We insert IPROBE to display i. After simulation, we obtain, 1(0) = 7.714 A from the display of IPROBE. When the switch is in position b, the schematic is as shown below. For inductor L1, we let IC = 7.714 A. By clicking Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step = 2 5. After Simulation, we click Trace/Add in the probe menu and display I(L1) as shown below. Note that i(oo) = 12 A , which is correct. 122 129T ————————————————————————————————————————————————————————— -— d a :: J__________L_-_______ Time We now know the initial and final values of the current through the inductor. i(O) = 7.714 A i(oo) =12 A To find the current through the inductor for any value of time, we need to know the time constant of the circuit. Using the circuit from part (b), Req=3l|6+4=2+4=6§2 t=L/Req =2/6=l/3s Therefore, i(t) = i(oo) + [ 1(0) — i(oo)] e"/‘ i(t) = 12 + [7.714 — 12] e31 = 12 — 4.286e'3‘ A 123 W0 APPLICATIONS Problem 7.20 [7.73] Figure 7.18 shows a circuit for setting the length of time voltage is applied to the electrodes of a welding machine. The time is taken as how long it takes the capacitor to charge from 0 to 8 V. What is the time range covered by the variable resistor? 100 kg to 1 M0 12V Figure 7.18 > Carefully DEFINE the problem. Each component is labeled completely. The problem is clear. > PRESENT everything you know about the problem. This is an RC circuit with a dc voltage source. When the welding machine is activated, the dc source supplies power to the RC circuit. To find the time in which it takes the capacitor to charge from 0 to 8 V, we need to find the response of the RC circuit, written as v0) = W») + [ v(0) — v<oo>l er where v(0) is the initial voltage across the capacitor, v(oo) is the steady—state value of the voltage across the capacitor, and T is the time constant of the RC circuit. > Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success. The three solution techniques that can be used are nodal analysis, mesh analysis, and basic circuit analysis. Basic circuit analysis can be used to solve this problem. > ATTEMPT a problem solution. v0) = v0») + [ v<0) — v(oo)] er V(0) = 0 V and V(oo) = 12 V W) = (12)(1-€"/’) Let the voltage at an unknown time, t0 , be equal to 8V. V(t0) = 8 = (12)(1 —e"°/‘) 124 _8_ “to/'r “to/1 1 . 12_1~e ———> e —3 t0 = T ln (3) For R=100 kQ, 1 = RC = (100x103)(2><10‘6)=0.2 s o =0.21n(3)=0.2197s For R=1MQ, , 1;:RC=(1X106)(2X10'6)=25 t0 =2ln(3)=2.197s Thus, 0.2197 s < to < 2.197 s > EVALUATE the solution and check for accuracy. For R = 100 kg), 1 = RC = (100x103)(2x10'6) = 0.2 s v(0.2197) = (12)(1— W) = (12)(1— e-0-2197/°~2) = 8 V . ForR=1MQ, I =Rc= (1x106)(2x10'6)=2s v(2.197) = (12)(1—6““) = (12)(1—6'2‘197/2) = s V Our check for accuracy was successful. > Has the problem been solved SATISFACTORILY? If so, present the solution; if not, then return to “ALTERNATIVE solutions” and continue through the process again. This problem has been solved satisfactorily. 0.2197 s < t0 < 2.197 s 125 ...
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This note was uploaded on 09/06/2010 for the course EE 50145 taught by Professor Chi during the Spring '10 term at UCSF.

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ProbSolv_Chapter07 - CHAPTER 7 - FIRST-ORDER CIRCUITS List...

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