A Textbook of Electrical Technology Volume 2 by B.L. THERAJA and A. K. THERAJA


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Simple Resistive Circuits [b] With R = 30 Ω we can calculate the current through R using current division, and then use this current to find the power dissipated by R, using the formula p = Ri2 : 40 + 80 iR = (20 A) = 16 A so pR = (30)(16)2 = 7680 W 40 + 80 + 30 [c] Write a KVL equation around the outer loop to solve for the voltage v, and then use the formula p = −vi to calculate the power delivered by the current source: −v + (60 Ω)(20 A) + (30 Ω)(16 A) = 0 Thus, so v = 1200 + 480 = 1680 V psource = −(1680 V)(20 A) = −33,600 W Thus, the current source generates 33,600 W of power.

5 Always work from the side of the circuit furthest from the source. Remember that the current in all series-connected circuits is the same, and that the voltage drop across all parallel-connected resistors is the same. [a] Circuit in Fig. 3(a): Req = [(7000 + 5000) 6000] + 8000 = 12,000 6000 + 8000 = 4000 + 8000 = 12 kΩ Circuit in Fig. 3(b): Req = [500 (800 + 1200)] + 300 + 200 = (500 2000) + 300 + 200 = 400 + 300 + 200 = 900 Ω Circuit in Fig. 3(c): Req = (35 + 15 + 25) (10 + 40) = 75 50 = 30 Ω Circuit in Fig.

Finally, KCL at each of the middle three nodes yields the currents in the two sources and the current in the middle 2 Ω resistor. These calculations are summarized in the figure below: p147(top) = −(147)(28) = −4116 W p147(bottom) = −(147)(21) = −3087 W Therefore the top source supplies 4116 W of power and the bottom source supplies 3087 W of power. , Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.

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