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Unformatted text preview: lst Order Circuit Probs 11/26/01 P81 © R. A. DeCarlo, P. M. Lin SOLUTIONS PROBLEMS CHAPTER 8 de (t) _ VC (2‘) de(t) vC (t) _ .
SOLUTION 8.1. (a) By KCL, C dt — R or dt + RC —0. Usmg 8.12 vC (t) 2 VC (me—m =106—t V where 17 = RC=1s. Plotting this from 0 to 5 sec
»t = 0:.0515;
>>vc = 10*exp(—t); >>plOt(t,VC) » grid »x1abe1('time in s’)
>>ylabel('vc(t) in V') vc(t) in V time in s (b) The solution has the same general form as for (a),iC(t) = iC (0+)et = —VC (0
R . lst Order Circuit Probs 11/26/01 P82 © R A D C
, . e arlo, P. M. Lin ic(t) in mA time in s (c) By linearity, if VC(0) is cut in half, all resulting responses are cut in half. If VC(0) is doubled, then all resulting responses are doubled. Alternately, one View this as a simple change of the initial condition
with the same conclusion reached from linearity. SOLUTION 8.2. (a) From inspection of the general form, 8.12, 0.1/1: = 1: 0.1/ RC :> C = 0.1/R = 5 pH
(b) Sincet = RC = 0.1, vC(t)=10e_10t V.
_ —t/Re C .
SOLUTION 8.3. (a) The general solution form is VC (1‘) = vC (0)6 m = vC(0)e ‘1 . Us1ng the
—(0.001)/‘t
0.001 18.394 v (0)6 0.00m
given data, take the following ratio, vC( )  — = 2.7183 = —£—————— — e . Hence, vC (0.002) ‘ 6.7668 wane—(comm ‘ »K = 18.394/6.7668 K 22.7183 »tau = le—3/log(K) tau =0.0010 »C = 566; >>Req = tau/C Req =200.0008 »% Req = R*4e3/(R+4e3) lst Order Circuit Probs 11/26/01 P83 © R. A. DeCarlo, P. M. Lin »R = Req*4e3/(4e3—Req) R =210.5272 »VCO = 6.7668/exp(—0.002/tau)
vCO: 49.9999 V 0)) >> % VC (I): 506
»t = 0:tau/100:5*tau;
me = VCO*exp(—t/tau);
»plot(t,vc) —1000t V >> grid
»xlabel('Time in ms’)
»ylabel('VC(t) in V') Capacitor Voltage (VII Time in ms SOLUTION 8.4. After one time constant the stored voltage, 8 V, decays to 8/e = 2.943 V. From the
graph, the time at which the output voltage is 2.94 V is approximately 0.19 5. Thus I = 0.19 s, and R = 'c/C = 190 Q. SOLUTION 8.5. (a) The general form of the inductor current is iL(t) = iL (0)e_t/T = 0.156.”1 A where 1 = L/R = 2 ><10~3 s. Plotting W) = owe—500‘ »t = 0:.01e—3210e3; »iL = O.15*exp(t/2e3);
»plOt(t,iL) »grid A from 0 to 10 msec yields: lst Order Circuit Probs 11/26/01 P8—4 © R. A. DeCarlo, P. M. Lin >>xlabe1('Time in s') >>ylabe1('Inductor Current') 0.1 6 l V
0.14 ............. ................................ ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, . ............ . . . ........................ . ..................... ,,,,,,,,, ,,
Q12 ......... ........ . . . . . , ................ ........................ .. . 0.1 i i 0.08
0.06 Inductor Current o_02 .. . . . , , .................................. . ............................................................... .. .................................................................................. .. 0 ‘ ‘ ' —
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 Time in s: (b) Here vLa) = —RiL(r> =—22.se“500t V, Inductor Voltage in V O 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 Timp. in s 225 —500z
e 3 —7.5e‘5°°’ V and for iL(0) = 250 H (c) Using linearity, for iL(0) = 50 mA, then vL (t) = 250 _
mA, then vL(t) = —22.5 1—56 e 500‘ = 37.Se ~500t V. lst Order Circuit Probs 11/26/01 P85 © R. A. DeCarlo, P. M. Lin __ . . 3
SOLUTION 8.7. (a) We desire to solve iL(t)= iL (0)6 m for 1L(0) and R 1n 1: =0.08 /(R +10 ). . . —(0.05m)/12 _
h . I m5) __ __ L = 3h = 7 Hence
Usmg the followmg ratlo, ————iL (0. 1 5 ms) — 1. 2 4 47 iL (0)6_(0.15m)/,C e . . »K = 9.197/1.2447
K =7.3889 >>tau = 0.1e3/log(K)
tau =5.0000605 »L = 0.08; >>Req = L/tau Req =1.6000e+03
»R = Req  1000 R = 599.9862 »iL0 = 9.197e—3/exp(~0.05e3/tau)
iLO = 0.0250 A .6
(b) I = 80m /(R +1000): 50” sec, My) = 0_0253—t/50><10 A_ [1.025 D
D
M . . . . . . . . . . . , . . . . . . . . . . . . . , . . . . . . . . . . . . . . _ . . . . . . . . . . . . . x . . . . . . . . . . . . . m . . . . . . . . . v . . Inductor Current (A) 0.25 0.15 0.2
Time in ms 0.3 SOLUTION 8.8. By Ohm’s law, vR(0+) = (4k 16k)iL(0+) = —32 V. The time constant
’c = L/(4k l6k)= ZSHsec, i.e., »Req = (4e3*16e3/(4e3+16e3)) Req =3200 »L = 0.08; lst Order Circuit Probs 11/26/01 P86 © R. A. DeCarlo, P. M. Lin >>tau = L/Req
tau =2.5000e05 Using the general equation, vR(t) = vR (0+)e—t"t = —32r2—t/25u V. Equivalently, 4
vR(t) =—ReqiL(t)= —Req0.01e—4X10 ‘ V. SOLUTION 8.9. (a) Note that the Thevenin resistance seen by the capacitor is Rth:
»R1 = 360+60*120/(60+120) R1 =400 »Rth = 400*1200/1600 Rth =300 Hence, vC (r) = vC (me—t" = 806m)” V Where 1: = RTHC = 300x 0.5 ><10’3 = 0.15 s. 01
O Capacitor Voltage (V)
.1:
O 3O 
1O
0 ' r
O 0.2 0.4 0.6 0.8 1
Time in s
(b) Here iC(0+) =—vC(0+)/RTH = —0.2667. Therefore iC(t)= iC(o+)e‘”t =—0.2667e"”°15 A.
Equivalently, ic(t)= ‘vC (t) = —0.2667e_t/0'15 A.
th (c) By VOltage diViSion’ VR(0+) = VC(0)( 60mm j: 8 V; for t > 0 vR(t) = 8e't/O'15 V. (60“ 120) +160 + 200 1st Order Circuit Probs 11/26/01 P87 © R. A. DeCarlo, P. M. Lin SOLUTION 8.10. First, ﬁnd the Thevenin equivalent seen at the left of the inductor. Introducing a test
source in place of the inductor we obtain the following KCL equation at that node. item. = vtest /lk +[vmt — 200(%&)]/200 . Let vtest = l V. Then >>itest = le—3 + (1 — 200*1e3)/200
itest =0.0050 »Rth = l/itest Rth =200 Thus Rm = 200 o, r = L/Rth = 0.25 ms, and iL(t) = 0.025e4000’ A. Next from 8.13b find vL(t) = —(RTH >< iL(0))e_4000t = —Se_4000tV, and from Ohm’s law ix(t) = 56490‘” mA_ SOLUTION 8.11. For all parts it is necessary to ﬁnd the Thevenin equivalent resistance seen by the
capacitor. To this end we apply an external test current to the remainder of the circuit to obtain:
Vtest : Rlitest + R2(itest + 0litest)
Thus
Rm = vtest lite” = R1 +R2(l+0t) = 120+ 70(1+ or) (a) With or = 4, Rm = 470 9,1 = RmC =0.1175 s, and W): vC (me—“1 = 506—851t V. 4s
O N
O (.0
O
I : : . Capacitor Voltage (V) _..
O 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Time in s (b) With 0‘ = 4’ Rm = 90 Q, I = RthC =0.0225 s, and vC (r) = 5034444t v_ lst Order Circuit Probs 1 1/26/01 P88 © R. A. DeCarlo, P. M. Lin apacitor Voltage (V C
N
o
o
o 0 0.02 0.04 0.06 0.08 0.1 0.12
Time in 5 Note how this is not a stable design as V increases exponentially without bound. (c) From the general equation developed at the beginning, Rm = 120 + 70(l+ a) > 0 requires that
oc > —2.7 143 . SOLUTION 8.12. Find the Thevenin resistance left of the inductor. Forcing a test current source into
the output node, vtest : Rlitest (itest _aRlitest) : + a100))itest = — 5000a)itest
and (a) Using the above equation, Rm = «850 Q and 1: = L/Rth = ~7.l429 ><10'5 3. Hence, W): 0.18400“ A, an unbounded response due to the presence of the negative equivalent resistance.
(b) Rth = 100 9,1 = L/Rth = 2.5 ><10_4 s, vR(0+)= —R2(l——0cR1)iL(0) =0 , but more importantly
vR2(t) = —R2(l—0tR1)iL(t) 2 0X iL(t) =0. (C) R1 + R2(1 —ocR1) = (150 — 06000) > 0 impliesoc < 0.03. SOLUTION 8.14. This is similar to problem 8.13. Here the current turns off at time zero instead of a switch opening. By current division i L (0+) = I s / 2 . The difference between this problem and problem 8.13 is that the Thevenin resistance seen by the inductor is different. Here, RTH = 2R IIO.5R = 0.4R. So for t > 0, iL = (IS /2)e_R‘ht/L = 0.51 se—Rt/Z'SL A. A sketch of this function plotted with respect to this new time constant will be identical to the one in problem 8.13. Deﬁne the time constant of problem 8.13 as told. The slower decay in the plot below represents the fall of the inductor current for problem 8.14 relative to that of problem 8.13 which is the faster decaying curve in the plot below.
13 EELS [14 1st Order Circuit Probs 11/26/01 P810 © R. A. DeCarlo, P. M. Lin SOLUTION 8.15. Over a long period of constant applied voltage, a capacitor looks like an open circuit.
3
By voltage division, vC (0+) = Z VS and 1: = 3RC. Hence —t/(3RC) V vC (r) = vC (0+)e‘”t = 0.75Vse ‘stﬂﬁ EDD
U'Im'hd {3
.p. . . . . . . . . . . . . . _ . . . . . . . . . . . . . . . . . . . . . . . . . , . . . . . . . . . . . . . . . . . . . . . . . . . . . .. El
m . . . . . . . . . . . . .  . . . . . . . . . . . . . . e . . . . . . . . Capacitor voltage {v}
D
Eu D
_ . . . . . . . . . . . . . _ . . . . . . . . . . . . . . ~ . . . . . . . . . . . . . . I . . . . . . . . . . . . . . . . . . . . . . . . . _ . . . . . . . . . . . . . D g t 21: 3 1:
=3”: =6RG =9Rc SOLUTION 8.16. Same problem as 8.15 for t < 0. For t > 0 the effective resistance changes. Rm = 3R / /R = 0.7512 and 1: = 0.75RC. Thus, VC (1‘) = 0.75Vse_t/(0'75RC) v. Same behavior as in the previous problem, except for a faster decay than in problem 8.15 due to a smaller effective resistance.
Note how the decreased resistance affects the RC and RL circuit differently. SOLUTION 8.17. For t < 0 the inductor looks like a short circuit. Let R1 = 1333/ /800= 500 Q. The
12 current supphed by the source IS I S = m = 0.02 A. By current lelSIOH,
1333
iL(0+) =0.02  m =0.0125 A For t > O, the switch is opened and the inductor sees only the 800 Q resistor. Hence, 1: = L/ R = 25 usec and 1st Order Circuit Probs 11/26/01 P811 © R. A. DeCarlo, P. M. Lin —40000! m A iL(t)—'= Inductor current in mA 0 20 4O 60 80 1 00
Time in micro seconds SOLUTION 8.18. (a) For t < 0 the applied voltage is constant and at t = 0, the capacitor is like an open 25k I .25k
CirCUit By voltage diViSiOn, VC (0+) = 30m = 25 V. For 0 < t < 1 ms, the source is off and the capacitor discharges through three resistors in parallel; thus I = = X X = ms and vC = 256—2000t —2000(0.00 1) (b) From continuity vC (0.001): 25e = 3.383. For t > lms, the capacitor keeps on discharging through only one resistance, the 25 k9 resistor; thus the new time constant is Tnew = 15 ms, and V0 (1‘) = 3.383e“("0001)/0.015 V. (c) PlotvC (t) = 25e“2000’[u(r) — u(t— 0.001)] + 3.383e‘(t“°'°°1)’°'°15 u(t — 0.001) V
»t = 0:01:12; >>vc = 25 *exp(2*t) .*(u(t)u(tl))+3.3 834*exp(—(tl)/ 15) .*u(t—l); >>plot(t,vc) >>ind >>xlabel('Tirne in milli—secs') »ylabel('Capacitor Voltage in V') lst Order Circuit Probs 11/26/01 P8—12 © R. A. DeCarlo, P. M. Lin 25 Capacitor Voltage in V TextEndiE 5L ................................................... .................................................................................. Time in milIisecs 54 . .
60 + 30 — 0.6 A. For t > O, the Thevenm res1stance seen by the inductor is Rth = (60+ 30) H720 = 80 Q and 1: = L/Rth = 1/160 s. Thus SOLUTION 8.19. (a) From Ohm’s law iL(0+) = 160t i (t) = i (0+ )e—m = 0.6e A. From Ohm’s law and current division
L L 720 . __ ~160t
———90+720le(:)_ 32e V v(t) = —60 X
(b) From continuity property, iL(0.01:L) = 0.6—160(0‘01) = 121.14 mA. For t > 10 ms, the Thevenin
resistance seen by the inductor is Rm = (690 + 30) II 720 = 360 Q and tnew = L/Ru1 2 1/720 s. Hence, iL(t) = 121.14e‘720U—0'01) mA for t > 10 ms. From Ohm’s law and current division 720 . = —720(t—0.01) _
—————720+720le(0 —41.793e u(t 0.01)V v(t) = —690 x Therefore iL(t)= 0.6e‘160‘[u(z)— u(t—0.01)]+0.121146—720(t—0'01)u(t—0.01) A lst Order Circuit Probs 11/26/01 P8—13 © R. A. DeCarlo, P. M. Lin 0.6 1
<
E 0.5  E 0.4 l
E 0,3 O
8 O 2 E ' TextEnd 0 5 10 15
Time in miniseconds SOLUTION 8.20. For both circuits we first compute the Thevenin resistance seen to the right of the
capacitor for 0 S t S 60 ms. If we excite the circuit to the right of the capacitor over this time interval, _ Vtest + (1—0'25)vtest then item — 200 100 = 0.0125vmt. Let Rthl = 00125 = 80 Q.
80
(a) For t < 0, the capacitor acts as an open circuit. Using voltage division, vC (0+) = m 80 = 30 V. For 0 S tS 60 ms, the time constant istl = R1.th = 40 ms, and
vC(t)= vC(0+)e“/T = 30;” (Rm IC) = 30a“25t v From continuity, vC (60Jr ms) = 6.694 V. The new Thevenin resistance is Rthz = 200 9. Thus for t > 60 ms, the time constant is’cz = 200C = 100 ms, and vC (t) = 6.694e—10(t_0’06) V. The resulting capacitor voltage is plotted below. lst Order Circuit Probs 11/26/01 P8—14 © R. A. DeCarlo, P. M. Lin 3O Capacitor Voltage (V)
61‘ ._
O
'—I' TextEnd O 50 100 150 200 250
time in ms (b) It is the same circuit as above for t < 0; thus vC (0+) = 30 V. However the Thevenin resistances seen
by the capacitor are different because there is no switch to disconnect the independent voltage source and its series resistance. First for 0 S t S 60 ms, the Thevenin resistance to the right remains as 1
0.0125 changes to Rth3 = Rthl // 133.3 2 50 9. Then new time constant is 1:3 2 Rth3C = 25 ms and for 0 S t S
60 ms Rthl = = 80 £2. However, for 0 S t S 60 ms, the Thevenin resistance seen by the capacitor vC (t) = 30640t v From continuity, vC (60+ ms) = 2.72 V. The new Thevenin resistance is Rth4 = 200/ / 133.3 2 80 (2. Thus for t > 60 ms, the time constant ist4 = 80C = 40 ms, and vC (t) = 2.72e—25(t_0'06) V. The resulting capacitor voltage is plotted below. 1st Order Circuit Probs 11/26/01 P815 © R. A. DeCarlo, P. M. Lin 30 25 N
O
_l' Capacitor Voltage (V)
61‘
I O 50 100 150 200 250
time in ms (0) For t < 60 ms, the voltage decays faster in (b) due to the smaller time constant. Similarly, for t > 60 ms. SOLUTION 8.21. Following, are the switching times with the time constants associated with them.
i=0 => Rth=20 kQ => 13:20 ms t=5 ms => Rth=4 k9 => 1:4 ms t=7.5 ms => Rm =800 Q => 1 =0.8 ms It follows that with t in ms, vC (t) = 10e_0‘050t[u(t) — u(t— 5)] + 7.788e‘025°("5)[u(t— 5) — u(t 7.5)] + 4.169e—1'25(t_7‘5)u(t — 7.5) V lst Order Circuit Probs 11/26/01 P816 © R. A. DeCarlo, P. M. Lin Capacitor Voltage (V) time in ms *SOLUTION TO 8.22. This solution is done in MATLAB. % Define switching times, inductance, and Thevenin equivalent resistances. t1: 1266;
t2: 1 866;
t3=Zle6;
L: 0.1;
rth1= 800;
rth2= 8000;
rth3=l600;
rth4= 32000; % Deﬁne time constants for each of the four time intervals. taul: L/rthl
tau2= L/rth2
tau3= L/rth3
tau4: L/rth4 taul = 1.2500e04
tau2= 1.2500e—05
tau3 = 6.2500e05
tau4 = ‘3.1250e—06 % Compute initial inductor currents for each of the four time intervals. ill: 100e—3;
i12=i11*exp(—tl/tau1) 1st Order Circuit Probs 11/26/01 P817 © R. A. DeCarlo, P. M. Lin il3=i12*exp(—(t2—tl)/tau2)
il4=il3 *exp(—(t3—t2)/tau3) i122 9.0846e02
il3 = 5.62146—02
i14= 5.3580e02 % Determine inductor currents for each of the four time intervals. Plot. t = 0:0.5e—7236e—6; seg1= i11*exp(—t/tau1) .*(ustep(t)ustep(t—t1)); seg2 =i12*exp(—(tt1)/tau2) .*(ustep(t—t1)ustep(t—t2));
seg3=il3*exp(—(t—t2)/tau3) .*(ustep(t—t2)ustep(t—t3));
seg4= i14*exp((t—t3)/tau4) .* ustep(t—t3); iL=segl + segZ +seg3 + seg4; plot(t,iL)
grid 1131] inductor CUH‘EI’lt in I'T'IF'.
U‘I
D El 5 1U 15 _ 2U 25 3D 35 4D
TIFﬂE In mints—sat: SOLUTION 8.23.
For circuits with a forced voltage, equation 8.190 is used as a general solution, t
vC (t) = vC (00) + [vC (t3) — vC (00)];[RTHC (a) At time zero the voltage is 0 V. As time approaches inﬁnity, the capacitor looks like an open with
voltage 10 V. The Thevenin resistance is 10 k9. Thus for t > 0. lst Order Circuit Probs 11/26/01 P818 © R. A. DeCarlo, P. M. Lin t vC(t)= 10+ [—10]e{2) =10(1— (3415’) V. t 2] =Se_0'5tV.
5 (b) With Vin(t) = 0 and VC (0*) = 5 V, vC (t) = 5A
10 f5; .a—h. % $3 '33 m E E : 3 D r :0: 5— 5.. E 4 3 2 U “E. g 2  31 u E: E! I] Time ins Time “15
Part (a) Part (b) (c) From linearity, vC (t) = 10(1— (3415’ ) + 5605’ = 10 — 5515’ V. Using Ohm’s law, Vin(t)__VC(t) . _ __ _ —0.5t _ —0.5t
104 104 ThUSlC(I)—1 (1 0.5e )_0.5e mA. icU‘): Capacitor Current (mA) 0 2 4 6 8 10
time in s (d) This is the same as (a), under the condition that the input is 1.5 times larger. Hence by linearity,
vC (t) = 1.5 x10(1— #51): 15(1— e—O'St) V.,
(6) By linearity, ANSWER = —2x(ANSWER to (b)) + 3x(ANSWER to (a)): 1st Order Circuit Probs 11/26/01 P8l9 © R. A. DeCarlo, P. M. Lin —O.5t vC (t) = —2>< 5605’ + 3x10(1— 605‘) =15(1— 605’): 30— 40e V SOLUTION 8.24.
(a) At t = 0—, the capacitor looks like an open circuit; therefore, by voltage division and the continuity _ 3R . . 3R
property, vC (0 )= vC (0+) = E 51 = 0.75Vsl. Similarly, at t = 00, vC (co) = Z} 52 = 0.75V32. The circuit time constant is 1: = (3R / /R)C = 0.75 RC. Hence _ t )
vC(t)= 0.75VS2 +0.75[VS1_Vs2]e (0.75m: . (b) A sketch will show an exponentially varying voltage from 0.75VS1 converging to 0.75VS2 with the
computed time constant.
(c) The response to the initial condition when the inputs are set to zero, zero—input response, is t vC (t) = VS 1e [OJSRC The zero order response, the response with 0V initial condition to a forced t
{075ch voltage, is vC (t) = V52 — Vs 26 . SOLUTION 8.25.
For RL circuits with a forced current, equation 8.1% is used as a general solution: iL(l)= iL(oo) +[iL(t;L)— iL(oo)
Since Rth = R = Q and L = H, we have T = 4 ms and
iL(t)= +[iL(l';)— iL(oo)]e_250(t_t0) (3) Here, i L (0) = 0 and as time approach inﬁnity, the inductor becomes a short and
iL(<>°) =10/100: 0.1 A. Thus iL(t)= 0.1(1— 525‘”) A, (b) Here iL (0) = 50 mA and because the input is zero, Moo) = 0. Thus, iL(t) = —0.05e‘250’ A. Plots
for parts (a) and (b) appear below. lst Order Circuit Probs 11/26/01 P8—20 © R. A. DeCarlo, P. M. Lin Inductor Current (A) 002} . . . . 0.06 = =
O 5 1O 15 20 Time in ms (0) By linearity 1m) = 0.1(1— 6250‘) — 0.05625“ = 0.1— 0.15625” A. Further, by KVL and Ohm's law, mt): vin(t)—100iL(t) implies mt): 10—10+15e*25°’ = 15e‘250t V. Inductor Voltage (V) extEnd O 5 10 15 20
Time in ms (d) Observe that the new initial condition is —().5 times the old one and that the new input voltage is 1.5
times the old one. Therefore, by linearity, and thus
VLU) = 15 — 15 +12.52_250t =12_5e—250t V The plot is similar to part (c) with initial point 12.5 instead of 15. SOLUTION 8.26. For this problem Rm = 2R / /0.5R = 0.4R in which case 1: = l/Rth = L/0.4R. lst Order Circuit Probs 11/26/01 P821 © R. A. DeCarlo, P. M. Lin (a) At t = 0, the inductor looks like a short circuit. Hence by current division, i L (0') = i L (0+) = 0.5151 . A similar argument yields i L (00) = 0.5152. Using the general form of the solution,
0.4Rt iL(t)= 0.51s2 +0.5[IS1 — 1,2]c L (b) A sketch will show an exponentially varying current from 0.5181 A converging to 0.5152. (c) The response to the initial condition when the inputs are set to zero, zeroinput response, is
0.4Rt i L(t) = 0.5151e L . The zero state response, the response with no initial condition, to the input
0.4Rt Iszu(t),is iL(t)=IS2 1—e L SOLUTION 8.27. For this problem, vC(0)= (20—1060“) =10 v and t=0
vC(o<>) =101s = 20 = 11m(20—10e*)4‘). Hence I, = 2 A. Further,
t—>°°
d t _
iC(t) = C VS; ) = 4Cﬂ'4‘ = 0.4e 0"” which implies that C = 0.1 F. Since 17 =1/0.4 =2.5 = (10+ R)C= 0.1(10+ R) , it follows that R = 15 Q. SOLUTION 8.28.
(a) The Thevenin resistance for this configuration is Rth = 1000/ / 1000 = 500 Q and 17 = RthC = 0.25 s. Hence vC (t) = vC (0+ )eﬂ‘l‘c =15e_4t V is the zeroinput response.
(b) Using a source transformation and voltage division, vC (co) = 3 V. Thus vC (t) = 3(1— e_4t) V. (0) Here vC (co) = 4 v, thus vC (t) = 4(1— 64”) v.
(d) This is the superposition of parts (b) and (c), i.e., vC (t) = 7(1— e4”) (e) The complete response is the superposition of parts ((1) and (a), i.e., VC (1) = 7 + 86‘” V.
(1) From linearity, vC (t) = 0.5 x 7(1— e”4‘)+ 2 ><15e‘4t = 3.5 + 265e—4t ...
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