DeCarloCh8Sol

# DeCarloCh8Sol - 1st Order Circuit Probs P8-1 R A DeCarlo P...

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1st Order Circuit Probs 11/26/01 P8-1 © R. A. DeCarlo, P. M. Lin SOLUTIONS PROBLEMS CHAPTER 8 S OLUTION 8.1 . (a) By KCL, C dv C ( t ) dt = - v C ( t ) R or dv C ( t ) dt + v C ( t ) RC = 0 . Using 8.12 v C ( t ) = v C (0) e - t / = 10 e - t V where = RC = 1 s . Plotting this from 0 to 5 sec »t = 0:.05:5; »vc = 10*exp(-t); »plot(t,vc) »grid »xlabel('time in s') »ylabel('vc(t) in V') 0 1 2 3 4 5 0 2 4 6 8 10 time in s vc(t) in V TextEnd (b) The solution has the same general form as for (a), i C ( t ) = i C (0 + ) e - t / = - 10 5 × 10 3 e - t = - 2 e - t mA = - v C ( t ) R .

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1st Order Circuit Probs 11/26/01 P8-2 © R. A. DeCarlo, P. M. Lin 0 1 2 3 4 5 -2 -1.5 -1 -0.5 0 time in s ic(t) in mA TextEnd (c) By linearity, if v C (0) is cut in half, all resulting responses are cut in half. If v C (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. S OLUTION 8.2 . (a) From inspection of the general form, 8.12, 0.1/ = 1 = 0.1/ RC C = 0.1/R = 5 μ F. (b) Since τ = RC = 0.1, v C ( t ) = 10 e - 10 t V. S OLUTION 8.3 . (a) The general solution form is v C ( t ) = v C (0) e - t / = v C (0) e - t / R eq C . Using the given data, take the following ratio, v C (0.001) v C (0.002) = 18.394 6.7668 = 2.7183 = v C (0) e - (0.001)/ v C (0) e - (0.002)/ = e 0.001/ . Hence, »K = 18.394/6.7668 K =2.7183 »tau = 1e-3/log(K) tau =0.0010 »C = 5e-6; »Req = tau/C Req =200.0008 »% Req = R*4e3/(R+4e3)
1st Order Circuit Probs 11/26/01 P8-3 © R. A. DeCarlo, P. M. Lin »R = Req*4e3/(4e3-Req) R =210.5272 »vC0 = 6.7668/exp(-0.002/tau) vC0= 49.9999 V (b) » % v C ( t ) = 50 e - 1000 t V »t = 0:tau/100:5*tau; »vc = vC0*exp(-t/tau); »plot(t,vc) »grid »xlabel('Time in ms') »ylabel('vC(t) in V') S OLUTION 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 s. Thus = 0.19 s , and R = τ /C = 190 . S OLUTION 8.5 . (a) The general form of the inductor current is i L ( t ) = i L (0) e - t / = 0.15 e - t / A where = L / R = 2 × 10 - 3 s. Plotting i L ( t ) = 0.15 e - 500 t A from 0 to 10 msec yields: »t = 0:.01e-3:10e-3; »iL = 0.15*exp(-t/2e-3); »plot(t,iL) »grid

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1st Order Circuit Probs 11/26/01 P8-4 © R. A. DeCarlo, P. M. Lin »xlabel('Time in s') »ylabel('Inductor Current') 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 Time in s Inductor Current TextEnd (b) Here v L ( t ) = - Ri L ( t ) = - 22.5 e - 500 t V. 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 -25 -20 -15 -10 -5 0 Time in s Inductor Voltage in V TextEnd (c) Using linearity, for i L (0) = 50 mA, then v L ( t ) = - 22.5 3 e - 500 t = - 7.5 e - 500 t V and for i L (0) = 250 mA, then v L ( t ) = - 22.5 250 150 e - 500 t = - 37.5 e - 500 t V. S OLUTION 8.6 . Since = L / R , we can solve for L = 5 mH. Then solving i L ( t ) = i L (0) e - t / for i L (0) at t = 4 μ sec yields i L (0) =15 mA.
1st Order Circuit Probs 11/26/01 P8-5 © R. A. DeCarlo, P. M. Lin S OLUTION 8.7 . (a) We desire to solve i L ( t ) = i L (0) e - t / for i L (0) and R in = 0.08 /( R + 10 3 ). Using the following ratio, i L (0.05 ms ) i L (0.15 ms ) = 9.197 1.2447 = i L (0) e - (0.05 m )/ i L (0) e - (0.15 m )/ = e 0.1 × 10 - 3 / = 7.3889. Hence »K = 9.197/1.2447 K =7.3889 »tau = 0.1e-3/log(K) tau =5.0000e-05 »L = 0.08; »Req = L/tau Req =1.6000e+03 »R = Req - 1000 R = 599.9862 »iL0 = 9.197e-3/exp(-0.05e-3/tau) iL0 = 0.0250 A (b) = 80 m /( R + 1000) = 50 sec, i L ( t ) = 0.025 e - t /50 × 10 - 6 A. S OLUTION 8.8 . By Ohm’s law, v R (0 + ) = - (4 k ||16 k ) i L (0 + ) = - 32 V. The time constant = L /(4 k ||6 k ) = 25 sec, i.e., »Req = (4e3*16e3/(4e3+16e3)) Req =3200 »L = 0.08;

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1st Order Circuit Probs 11/26/01 P8-6 © R. A. DeCarlo, P. M. Lin »tau = L/Req tau =2.5000e-05 Using the general equation, v R ( t ) = v R (0 + ) e - t / = - 32 e - t /25 V. Equivalently, v R ( t ) = - R eq i L ( t ) = - R eq 0.01 e - 4 × 10 4 t V.
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