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Download Circuit Analysis II with MATLAB Applications by Steven T. Karris PDF

By Steven T. Karris

Designed to be used in a moment path in circuit research, this article engages a whole spectrum of circuit research similar topics starting from the main summary to the main sensible. this isn't a math textual content. in spite of the fact that, the differential equations and Laplace transformation fabric awarded during this textual content is sufficient for the derivation of pertinent kinfolk and there's no have to check with complicated math texts. Featured are equipment of expressing indications by way of the simple services, an creation to moment order circuits, and a number of other examples of reading electrical circuits utilizing Laplace transformation equipment. notwithstanding no longer written explicitly for use with MATLAB, this article offers many beneficial information and techniques for MATLAB, permitting scholars to get the main out of the preferred software. all the info supplied is designed to be coated in a single semester or quarters. for more information. please stopover at the Orchard courses web site.

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Extra resources for Circuit Analysis II with MATLAB Applications

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887 150 50 -50 0 3 6 9 12 9 12 iL(t) 10 0 3 6 -10 3.  ` At t = 0 the circuit is as shown below. 100 : +  20 H  iL 0 +  v 0  C 400 : 100 V 1 e 120 F At this time the inductor behaves as a short and the capacitor as an open. 2 A . 2 = V 0 = 80 V and this establishes the first initial condition as V 0 = 80 V . For t ! 0 the circuit is as shown below. 2 - = 24 (4) = -------------- = --------------1 e 120 C Equating (3) and (4) we get – 2k 1 – 3k 2 = 24 (5) and simultaneous solution of (2) and (5) yields k 1 = – 36 and k 2 = 16 .

887 150 50 -50 0 3 6 9 12 9 12 iL(t) 10 0 3 6 -10 3.  ` At t = 0 the circuit is as shown below. 100 : +  20 H  iL 0 +  v 0  C 400 : 100 V 1 e 120 F At this time the inductor behaves as a short and the capacitor as an open. 2 A . 2 = V 0 = 80 V and this establishes the first initial condition as V 0 = 80 V . For t ! 0 the circuit is as shown below. 2 - = 24 (4) = -------------- = --------------1 e 120 C Equating (3) and (4) we get – 2k 1 – 3k 2 = 24 (5) and simultaneous solution of (2) and (5) yields k 1 = – 36 and k 2 = 16 .

7 Response of Parallel GLC Circuits with AC Excitation The total response of a parallel GLC (or RLC) circuit that is excited by a sinusoidal source also consists of the natural and forced response components. The natural response will be overdamped, critically damped, or underdamped. The forced component will be a sinusoid of the same frequency as that of the excitation, and since it represents the AC steady-state condition, we can use phasor analysis to find the forced response. We will derive the total response of a parallel GLC (or RLC) circuit which is excited by an AC source with the following example.

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