By Syed Nasar
Schaum’s strong problem-solver delivers 3,000 difficulties in electrical circuits, absolutely solved step by step! The originator of the solved-problem consultant, and scholars’ favourite with over 30 million research publications offered, Schaum’s bargains a diagram-packed timesaver that can assist you grasp all sorts of challenge you’ll face on exams.
difficulties disguise each region of electrical circuits, from easy devices to complicated multi-phase circuits, two-port networks, and using Laplace transforms. pass on to the solutions and diagrams you wish with our specified, cross-referenced index. suitable with any school room textual content, Schaum’s 3000 Solved difficulties in electrical Circuits is so whole it’s the correct device for graduate or specialist examination prep!
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Additional info for 3000 Solved Problems in Electrical Circuits
The average power in the 5-11 resistor is 20 W. I 1= o ~-5 = (b) -+2A The direction of I through the 5-0 resistor is determmed by noting that the polarity of the 90-V source requires that the current pass from d to c. Thus d is positive with respect to c and v'ie = (2)(5) = 10 V. (a) An ideal voltmeter indicates the voltage without drawing any current. infinite resistance. KVL applied to the clo~ed·lo()p acdba results in I) - 10 + 0 - VM = It may be considered as having an 0 VM= -lOV If the meter is of the digital type, it will indicate -10 V.
I In this case the equivalent resistance R = 1 + 3(2 + 4) = 3 0 3+2+4 Power drawn from the battery is 15 x 5 = 75 W. 81 A resistive circuit is shown in Fig. 3-27a. R = 15 = 5 A 3 Determine the equivalent resistance R. I The circuit reductions are shown in Fig. ::': thus 3-27b through d, from which R = 1 + 2 = 3 O. fl. I I. 17... - 14- VIfI ~ (,f2.. (L.. >+£ - 2 1,.. J ------------------------~ (ri) Fig. 82 D CHAPTER 3 Calculate the current through the 3-0 resistor and the voltage across the 1-0 resistor of the circuit of Fig.
N.. f\. (d) Fig. 3-16 By adding the powers absorbed by the resistors of Fig. 3-16e, verify that the sum is equal to the power supplied by the 36-V source. I From the results of Prob. 50, Thus, Is 0 = 4 A and 120 = 8 A Vso=4 x 5=20V and V20 = 8 x 2 = 16 V Also the volt ages across the top and bottom of the 3-0 resistors become 16 and 20 V respectively as shown. The respective currents in these resistors are ¥ A and ~ A. Hence the current I (Fig. 3-16a) in the vertically I = ¥ - 4 = ~ A. 52 Find the resistance between the terminals ad for the interconnected resistors shown in Fig.