Electric Circuits (9th Edition) Answer Manual by James Nilsson, Susan Riedel

By James Nilsson, Susan Riedel

Resolution handbook purely (complete, aka even and odds)

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Extra resources for Electric Circuits (9th Edition) Answer Manual

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15 [a] Plot the v − i characteristic From the plot: R= ∆v (130 − 50) = = 8Ω ∆i (10 − 0) When it = 0, vt = 50 V; therefore the ideal voltage source has a voltage of 50 V. 25A 8 Note that this result can also be obtained by extrapolating the v − i characteristic to vt = 0. , 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.

8 200 Ω 30 6 p = Ri2 180 Dep. 8 mW + −725 mW = −741 mW Thus, the total power generated in the circuit is 741 mW. 5 mW = 741 mW Thus, the total power absorbed in the circuit is 741 mW and the power in the circuit balances. , 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.

3–2 CHAPTER 3. Simple Resistive Circuits [a] With the simplified circuit we can use Ohm’s law to find the voltage across both the current source and the 12 Ω equivalent resistor: v = (12 Ω)(5 A) = 60 V [b] Now that we know the value of the voltage drop across the current source, we can use the formula p = −vi to find the power associated with the source: p = −(60 V)(5 A) = −300 W Thus, the source delivers 300 W of power to the circuit. [c] We now can return to the original circuit, shown in the first figure.

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