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

By J. David Irwin, R. Mark Nelms

Irwin's easy Engineering Circuit research has equipped an outstanding recognition for its hugely available presentation, transparent reasons, and vast array of worthy studying aids. Now in a brand new 8th variation, this hugely obtainable ebook has been fine-tuned and revised, making it more desirable or even more uncomplicated to take advantage of. It covers such issues as resistive circuits, nodal and loop research options, capacitance and inductance, AC steady-state research, polyphase circuits, the Laplace rework, two-port networks, and lots more and plenty extra.

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4 Find the equivalent inductance of the network in Fig. 4 at the terminals A-B. All inductors are 12mH. 43 A B Fig. 1 The equations for the waveforms in the 4 two millisecond time intervals are listed below. v (t ) = mt + b 2 = t 2 × 10 −3 =2 2 = −2+ t 2 × 10 −3 4 = + 16 − t 2 × 10 −3 =0 0 ≤ t ≤ 2ms 2 ≤ t ≤ 4ms 4 ≤ t ≤ 6ms 6 ≤ t ≤ 8ms t < 0, t > 8ms Note that within each interval we have simply written the equation of a straight line using the expression y = mx + b or equivalently v(t) = mt + b where m is the slope of the line and b is the point at which the line intersects the v(t) axis.

3(a) Before we begin our analysis, we note that resistors R3 and R4 are in parallel and so we first reduce the network to that shown in Fig. 3(b). + C R2 + R1 v1 v C 6k v 0 (t ) 50µF 6k + R = 3k t=0 + - 12 Fig. 3(b) Now that the network has been simplified, we begin our analysis +v R 1 = 6 k v1 R 2 = 6k - C + R = 3k 12V + - Fig. 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.

In addition, L4 and L6 are in parallel. Therefore, if we combine elements so that L25 = L2⎟ ⎜L5 and L46 = L4⎟ ⎜L6, then the circuit can be reduced to that in Fig. 4(c). L 25 L1 C A L3 L 46 B Fig. 4(c) However, we note now if we did not see it earlier that L25 is in parallel with L46 so that the network can be reduced to that shown in Fig. 4(d). L1 C A L3 L 2456 B Fig. 4(d) Where L2456 = L25⎟ ⎜L46. 1 Use the differential equation approach to find i 0 (t ) for t > 0 in the circuit in Fig. 1 and plot the response including the time interval just prior to opening the switch.

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