By Steven Karris
This article is written to be used in a moment direction in circuit research. It features a spectrum of matters starting from the main summary to the main functional, and the cloth should be lined in a single semester or quarters.The reader of this publication must have the normal undergraduate wisdom of an introductory circuit research fabric equivalent to Circuit research I with MATLAB Computing and Simulink / SimPowerSystems Modeling, ISBN 978-1-934404-17-1. one other prerequisite will be a simple wisdom of differential equations, and typically, engineering scholars at this point have taken all required arithmetic classes. Appendix H serves as a evaluation of differential equations with emphasis on engineering comparable subject matters and it is suggested for readers who may have a overview of this topic. for more information. please stopover at the Orchard courses web site.
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Extra info for Circuit Analysis II with MATLAB Computing and Simulink SimPowerSystems Modeling
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.