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 consultant, and scholars’ favourite with over 30 million research courses offered, Schaum’s deals a diagram-packed timesaver that can assist you grasp all kinds of challenge you’ll face on checks.
difficulties conceal each zone of electrical circuits, from uncomplicated 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 particular, 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 device for graduate or expert examination prep!
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Extra resources for 3000 Solved Problems in Electrical Circuits
Power delivered by the 12-V source is 12 x 1 = 12 W. 0 = - 3 W. Negative sign is used since the current is going inlO the source. Thus the dependent source is absorbing (rather than delivering) power. 67 D 31 A 500-0 resistor is connected in parallel with a 250-0 resistor and the combination is fed by a 25-A current source. Calculate the power absorbed by each resistor. I The circuit is shown in Fig. 3-23. 695 kW I 500£2.. Fig. 68 Determine the voltage across the resistors of Fig. 3-23. same as the total power dissipated in the resistors.
4-l4? How much power is consumed in the I Power = 1~(10) W. From Eqs. (1) and (2) of Prob. 22 13 = 0 and and power = O. Determine the current I supplied by the battery to the resistive network shown in Fig. 4-15. Icv Fig. 23 D CHAPTER 4 For the network shown in Fig. 4-16, calculate the power supplied (or absorbed) by each voltage source. l. _~ Fig. 24 Determine the power absorbed by each resistor of the network of Fig. 4-16. Verify that the sum of the powers absorbed by the resistors is equal to the total power supplied by the sources.
42 110 n 50 - V 10 50 - 60 10 = ___2 = - - - = -1 A In the circuit of Fig. 4-23a calculate the current through the 50-V voltage source and the voltage across the 5-A current source. 1 = k(200 - 3V,) = k(200 - 3 x 60) = ~ A 50-V voltage source. From Eq. (3) 2V3 = V2 - 100 = 60 - 100 = -40 or V3 = -20 V. Voltage across the 5-A source = V2 - V3 = 60 - (-20) = 80 V. I From Eq. (1) of Prob. 43 For the circuit of Fig. 4-23a verify that the total power dissipated in the resistors equals the total power supplied by the two sources.
3000 Solved Problems in Electrical Circuits by Syed Nasar