By Syed Nasar
Schaum’s robust problem-solver offers 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 publications bought, Schaum’s bargains a diagram-packed timesaver that will help you grasp all sorts of challenge you’ll face on assessments.
difficulties hide each region of electrical circuits, from easy devices to advanced multi-phase circuits, two-port networks, and using Laplace transforms. move on to the solutions and diagrams you would like with our specific, cross-referenced index. appropriate with any school room textual content, Schaum’s 3000 Solved difficulties in electrical Circuits is so entire it’s definitely the right software for graduate or expert examination prep!
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Extra resources for 3000 Solved Problems in Electrical Circuits
Jb Fig. 6 10' X = --~ - =30 - 11 ==30A 10 == 20 A A three-wire dc line supplying a resistive bank of loads is shown in Fig. 3·33. terminals a and e is 240 V, determine the voltage between a and b. 8-!! 7 V. BJl 'VV'v 24 n lOA 10 = 6 67 A . 7 V. 10. fl.. f'l.. 92 Refer to the circuit of Fig. 3-33. I Fig. 3-33 If 240 V is now applied across the lines ab, what is the voltage across be? 60V. 13V. 53 V. 93 37 In the circuit of Fig. 3-33, with line a open, determine the resistance between the terminals be.
77, Verify that the sum of these powers equals the Ptotal = 90W = VI = 15 x 6 = 90 W . If the 3-0 resistor in the circuit of Fig. 3-26a is short-circuited, how much power will now be drawn from the battery? Also, determine the voltage across the 2-0 resistor. I From Fig. 3-26a it follows that short-circuiting the 3-0 resistor short-circuits the 6-0 resistor as well as the series combination of the 2-0 and 4-0 resistors. Hence, V2 n = 0 V. The current is limited by the 1-0 resistor, and we have 1= 1[- = 15 A.
How much current flows through the 1-n resistor of the network of Fig. 4-21a? I From Fig. l = 10 - 5 x 2 = 0 or Find the value of R in Fig. 4-22 such that the power supplied by the 100-V source to the network is the same as the power supplied by the 5-A source. Fig. :::: __ V-lOO R 10 I At node 1: V-lOO ------w-+1 At node 2: _ 100 _ 5 4 - For equal power: (1) 20 - (2) LOCI/ I = 5 V (3) From Eqs. (1) and (2) we obtain: \' V R +10 = 15 (4) V 14 +10 = 15 (5) V Thus, R J. = (6) Finally, Eqs. 40 R=20n Find the current in the lO-n resistor of the circuit shown in Fig.
3000 Solved Problems in Electrical Circuits by Syed Nasar