By J. David Irwin, Robert M. Nelms
Maintaining its available method of circuit research, the 10th version contains much more positive aspects to have interaction and inspire engineers. fascinating bankruptcy openers and accompanying pictures are incorporated to reinforce visible studying. The text introduces figures with color-coding to noticeably enhance comprehension. New difficulties and multiplied software examples in PSPICE, MATLAB, and LabView are incorporated. New quizzes also are further to aid engineers make stronger the main innovations.
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Additional info for Basic Engineering Circuit Analysis (10th Edition)
SOLUTION – 1 ± – + 2 4V 11 A ± – 8Ix 8A First, we must determine the power absorbed by each element in the network. 5 Find the power that is absorbed or supplied by the circuit elements in the network in Fig. 5. 5 ± – ± – Ix=4 A 4Ix ANSWER: P24 V = 96 W supplied; P1 = 32 W absorbed; P4Ix = 64 W absorbed. 5 W absorbed. 2 Ix or supplied by the network elements in Fig. 6. 5 A 2A + Ix 15 V 3 – + 25 V ANSWER: Ix = -2 A. 7 Find Ix in Fig. 7 using Tellegen’s theorem. 7 The charge that enters the BOX is shown in Fig.
16b and c; hence, when we employ these elements a short time later, we must describe the units of the factors r and g. 16 Four different types of dependent sources. 3 Given the two networks shown in Fig. 17, we wish to determine the outputs. + 11 EXAMPLE In Fig. 17a the output voltage is Vo = VS or Vo = 20 VS=(20)(2 V)=40 V. Note that the output voltage has been amplified from 2 V at the input terminals to 40 V at the output terminals; that is, the circuit is a voltage amplifier with an amplification factor of 20.
Since the power absorbed is a function of the square of the current, the power absorbed in the two cases is quite different. Let us now demonstrate the practical utility of this simple voltage-divider network. Consider the circuit in Fig. 17a, which is an approximation of a high-voltage dc transmission facility.