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Extra info for Circuit Analysis II with MATLAB Computing and Simulink SimPowerSystems Modeling

Example text

50) and the constants k 1 and k 2 will be evaluated from the initial conditions. 51) k1 + k2 = 5 The second equation that is needed for the computation of the values of k 1 and k 2 is found from dv dt dv dt the other initial condition, that is, i L  0  = 2 A . 55) Check with MATLAB: syms t % Define symbolic variable t. 17. 17. , and the maximum voltage is approximately 24 V . 55), set it equal to zero, and solve for t . 76 V A useful quantity, especially in electronic circuit analysis, is the settling time, denoted as t S , and it is defined as the time required for the voltage to drop to 1% of its maximum value.

62) v  t  = e  5 + k2 t  As before, we need to compute the derivative dv  dt in order to apply the second initial condition and find the value of the constant k 2 . 19. 19. 67), we see that at t = 0 , v  t  = 5 V and thus the second initial condition is satisfied. 67). 20, i L  0  = 2 A and v C  0  = 5 V . Compute and sketch v  t  for t  0 . 20. 5 Solution: This is the same circuit as the that of the two previous examples except that the resistance has been increased to 50  . 96 and as before, 2 1 1  0 = -------- = ---------------------------- = 64 10  1  640 LC Also,  20   2P .

2 A 6 6 At t = 0 + the circuit is as shown below. 4 (3) t=0 Also, --------C- = ---C- and at t = 0 dv C --------dt t=0 iC  0  0 = ----------- = ---- = 0 (4) C C because at t = 0 the capacitor is an open circuit. 2e  2t + 1  We find i L  t  from i R  t  + i C  t  + i L  t  = 0 or i L  t  = – i C  t  – i R  t  where i C  t  = C  dv C  dt  and i R  t  = v R  t    1 + 2  = v C  t   3 . 4e  t + 1  12 3 6.    At t = 0 the circuit is as shown below where i L  0  = 12  2 = 6 A , v C  0  = 12 V , and thus the initial conditions have been established.