By Allan R. Hambley
<P style="MARGIN: 0px"> For undergraduate introductory or survey classes in electric engineering.
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<P style="MARGIN: 0px"> ELECTRICAL ENGINEERING: rules AND purposes, 5/e is helping scholars research electrical-engineering basics with minimum frustration. Its targets are to give uncomplicated recommendations in a basic atmosphere, to teach scholars how the rules of electric engineering follow to express difficulties of their personal fields, and to augment the general studying approach. Circuit research, electronic structures, electronics, and electromechanics are coated. a large choice of pedagogical good points stimulate pupil curiosity and engender knowledge of the material’s relevance to their selected profession.
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Additional info for Electrical Engineering: Principles and Applications 5th - Solutions
26 Under open-circuit conditions, 5 A circulates clockwise through the current source and the 10-Ω resistance. The voltage across the 10-Ω resistance is 50 V. No current flows through the 40-Ω resistance so the open circuit voltage is Vt = 50 V. With the output shorted, the 5 A divides between the two resistances in parallel. The short-circuit current is the current through the 40-Ω 10 = 1 A. Then, the Thévenin resistance is resistance, which is isc = 5 10 + 40 Rt = v oc / isc = 50 Ω. 27 Choose the reference node at the bottom of the circuit as shown: Notice that the node voltage is the open-circuit voltage.
261 V. If the source is reversed, the algebraic signs are reversed in the I matrix and consequently, the node voltages are reversed in sign. 52 To minimize the number of unknowns, we select the reference node at one end of the voltage source. Then, we define the node voltages and write a KCL equation at each node. 0588 A . The 20-Ω resistance does not appear in the network equations and has no effect on the answer. The voltage at the top end of the 10-Ω resistance is 15 V regardless of the value of the 20-Ω resistance.
8125 µA. 43 V . 2857 A. 49* Writing a KVL equation, we have v 1 − v 2 = 10 . 333 . 50 v2 − v1 5 − 5 v1 5 + v2 10 =1. 333 A . 261 V. If the source is reversed, the algebraic signs are reversed in the I matrix and consequently, the node voltages are reversed in sign. 52 To minimize the number of unknowns, we select the reference node at one end of the voltage source. Then, we define the node voltages and write a KCL equation at each node. 0588 A . The 20-Ω resistance does not appear in the network equations and has no effect on the answer.
Electrical Engineering: Principles and Applications 5th - Solutions by Allan R. Hambley