By Syed Nasar

Schaum’s robust problem-solver provides 3,000 difficulties in electrical circuits, totally solved step by step! The originator of the solved-problem advisor, and scholars’ favourite with over 30 million learn courses bought, Schaum’s deals a diagram-packed timesaver that will help you grasp every kind of challenge you’ll face on checks.

difficulties hide each quarter of electrical circuits, from simple devices to complicated multi-phase circuits, two-port networks, and using Laplace transforms. pass on to the solutions and diagrams you would like with our specified, cross-referenced index. appropriate with any school room textual content, Schaum’s 3000 Solved difficulties in electrical Circuits is so whole it’s the ideal software for graduate or specialist examination prep!

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**Additional resources for 3000 Solved Problems in Electrical Circuits**

**Example text**

TL-. , " '-I _\'.... - . , t-: ~ [e) Fig. 47 Solve for the power supplied by the 12-V source of Fig. 4-25a using the following steps: First convert the two current sources to voltage sources. Then use network reduction and, finally, obtain the power by applying KVL and KCL. Verify that the result is the same as in Prob. 46. I The source transformations and the network reduction are shown in Fig. 4-26a. Thus we obtain the circuit shown in Fig. 4-26b. 56 W. KIRCHHOFF'S LAWS 0 57 Fig. 48 Without using network reduction and source transformation, solve Prob.

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,.

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.