Circuits

Basic Engineering Circuit Analysis, Problem Solving by J. David Irwin, R. Mark Nelms

By J. David Irwin, R. Mark Nelms

Irwin's simple Engineering Circuit research has outfitted an effective attractiveness for its hugely available presentation, transparent motives, and huge array of priceless studying aids. Now in a brand new 8th variation, this hugely available ebook has been fine-tuned and revised, making it more beneficial or even more straightforward to take advantage of. It covers such issues as resistive circuits, nodal and loop research ideas, capacitance and inductance, AC steady-state research, polyphase circuits, the Laplace remodel, two-port networks, and masses extra.

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3(c) 57 + R 2 = 6k v (0 + ) 0 - + 6V R 1 = 6k R = 3k + 12V - Fig. 3(d) 6k 6k R TH 3k Fig. 3(e) Step-1 v 0 (t ) = k 1 + k 2 e −t τ Step-2 In steady-state prior to switch action, the capacitor looks like an open-circuit and the 12-V source is directly across the resistor R = 3kΩ. As shown in Fig. 3(c) the voltage v1 across R1 is equal and opposite to vC. Since the voltage of the 12-V source is divided between R1 and R2 we can use voltage division to find v1 as ⎛ R1 v 1 = 12 ⎜⎜ ⎝ R1 + R 2 ⎞ ⎟⎟ = 6V ⎠ hence v C (0 − ) = − v1 = − 6V = v C (0 + ) Step-3 The new circuit, valid only for t = 0 + is shown in Fig.

2(b). 55 i (t ) R 1 = 1k i = 1mH R 2 = 3k Fig. 2(b) If we let R = R1⎟ ⎜R2 then the differential equation for the inductor current is L di (t ) + R i (t ) = 0 dt The solution of this equation is of the form −t i (t ) = k 1 + k 2 e τ The differential equation has no constant forcing function and hence k1 = 0. , L = R= 3 Ω . This equation produces a τ value of 4k τ= 4 µ sec . 5 × 10 5 t mA, t > 0 The circuit is redrawn for convenience in Fig. 3(a). C 50µF R1 6k R2 6k + v 0 (t ) - R 4 6k R3 6k t=0 + 12V - Fig.

2 −8 16 +8= V 3 3 Recall that when employing source transformation, at a pair of terminals we can exchange a voltage source VS in series with a resistor RS for a current source Ip in parallel with a resistor Rp and vice versa, provided that the following relationships among the parameters exist. VS RS Rp = RS Ip = Now the original circuit is shown in Fig. 2(a). 8kΩ 6kΩ 12V + + V0 4kΩ 6mA - Fig. 2(a) Note that we have a 12V source in series with a 6kΩ resistor that can be exchanged for a current source in parallel with the resistor.

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