By J. David Irwin, R. Mark Nelms

ISBN-10: 0471740268

ISBN-13: 9780471740261

Irwin's uncomplicated Engineering Circuit research has equipped a high-quality acceptance for its hugely obtainable presentation, transparent factors, and vast array of invaluable studying aids. Now in a brand new 8th version, this hugely obtainable publication has been fine-tuned and revised, making it more suitable or even more uncomplicated to exploit. It covers such themes as resistive circuits, nodal and loop research options, capacitance and inductance, AC steady-state research, polyphase circuits, the Laplace remodel, two-port networks, and lots more and plenty extra.

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**Extra resources for Basic Engineering Circuit Analysis, Problem Solving Companion**

**Example text**

T=0 5 IS = A k 5kΩ R 1 = 1k i (t ) 1mA R 2 = 3k Fig. 2(a) In the steady-state time interval prior to switch action, the inductor looks like a shortcircuit. Therefore, in this time period t < 0, the initial inductor current is iL(0-) = IS = 5mA At t = 0 the switch changes positions and hence for t > 0 the network reduces to that 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.

4 Use the step-by-step method to find v 0 (t ) for t > 0 in the network in Fig. 4. 51 1 H 3 t=0 4Ω + 2Ω 12V +- v 0 (t ) 2Ω 12V + - Fig. 5 Given the network in Fig. 05F Fig. 6 Find i 0 (t ) for t > 0 in the circuit in Fig. 6 and plot the response including the time interval just prior to closing the switch. 1 F 120 t=0 12V + - i 0 (t ) 24Ω 24Ω Fig. 1 We begin our solution by redrawing the network and labeling all the components as shown in Fig. 1(a) R 3 i X (t ) -+ t=0 12V R1 R4 R5 C R2 i 0 (t ) Fig.

4(e) A simple voltage divider indicates that the output voltage is ⎛ 2 ⎞ ⎟⎟ = − 6V v 0 (∞ ) = − 12 ⎜⎜ ⎝2 + 2⎠ Step-5 The Thevenin equivalent resistance obtained by looking into the circuit from the terminals of the inductor with all sources made zero is derived from the network in Fig. 4(f). R TH 4Ω 2Ω 2Ω Fig. 4(f) Clearly RTH = 4⎥⎜(2 + 2) = 2Ω Then the time constant is 1 L 3 1 τ = = = sec . 5 (a) Applying KVL to the closed path yields v S (t ) = Ri (t ) + di(t ) 1 t ∫t i (x ) dx + L dt C 0 differentiating both sides of the equation we obtain dv S (t ) di (t ) i (t ) d 2 i (t ) =R + +L dt dt C dt 2 By rearranging the terms, the equation can be expressed in the form L or d 2 i (t ) di (t ) i (t ) dv S (t ) +R + = 2 dt dt C dt d 2 i (t ) R di (t ) 1 1 dv S (t ) i (t ) = + + 2 dt L dt RC L dt Using the circuit component values yields d 2 i (t ) di (t ) 1 dv S (t ) ( ) 7 10 i t + + = dt 2 dt 2 dt (b) The characteristic equation for the network is s 2 + 7s + 10 = 0 (c) The network’s natural frequencies are the roots of the characteristic equation.

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

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