DeCarloCh8Sol

DeCarloCh8Sol - 1st Order Circuit Probs 11/26/01 P8-1 R. A....

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1st Order Circuit Probs 11/26/01 P8-1 © R. A. DeCarlo, P. M. Lin SOLUTIONS PROBLEMS CHAPTER 8 SOLUTION 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. SOLUTION 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. SOLUTION 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)
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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') 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 s. Thus = 0.19 s , and R = τ /C = 190 . SOLUTION 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.
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DeCarloCh8Sol - 1st Order Circuit Probs 11/26/01 P8-1 R. A....

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