By Syed Nasar
Schaum’s strong problem-solver promises 3,000 difficulties in electrical circuits, absolutely solved step by step! The originator of the solved-problem advisor, and scholars’ favourite with over 30 million learn publications offered, Schaum’s deals a diagram-packed timesaver that will help you grasp every kind of challenge you’ll face on assessments.
difficulties hide each zone of electrical circuits, from simple devices to advanced multi-phase circuits, two-port networks, and using Laplace transforms. cross on to the solutions and diagrams you would like with our certain, cross-referenced index. suitable with any lecture room textual content, Schaum’s 3000 Solved difficulties in electrical Circuits is so whole it’s the precise software for graduate or expert examination prep!
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Extra resources for 3000 Solved Problems in Electrical Circuits
I Determine the currents I1 and 12 To apply mesh analysis, we first transform the to-A current source in parallel with the 5-0 resistor to a 10 x 5 = 50 V voltage source in series with a 5-0 resistor. Thus, we obtain the network of Fig. 85 A. Solve for I1 and 12 of the network of Fig. 4-lOa by nodal analysis. KIRCHHOFF'S LAWS IOn D 47 IOn sn + 2() lOA n o (a) (b) Fig. 4-10 I For nodal analysis, define the node volt ages VI and V2 as shown in Fig. 4-lOa. 85A For the network of Fig. 4-lOa, calculate the power supplied by the current source and by the voltage source.
3-36 (b) From Fig. 09 kO A dc generator having an internal resistance of 1 0 supplies a resistive load shown in Fig. 3-37 a. value of R x will the load draw the maximum power from the generator? I For what First we convert the delta-connected resistors to an equivalent wye, shown in Fig. 3-37 b, which is finally reduced to the circuit of Fig. 3-37c. Therefore, R load lOR x lOR x = R ab = 10 + R x + 10 + R x For maximum power transfer (see Prob. 3-28), or 20R x +R = 10 x 40 D CHAPTER 3 ~ tt l' R,.
The current is limited by the 1-0 resistor, and we have 1= 1[- = 15 A. Power drawn from the battery is VI = 15 x 15 = 225 W. 80 How much power is drawn from the battery if the 6-0 resistor of the circuit of Fig. 3-26a is open-circuited? 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.
3000 Solved Problems in Electrical Circuits by Syed Nasar