Electric Circuits, 9th Edition P15

Electric Circuits, 9th Edition P15. Designed for use in a one or two-semester Introductory Circuit Analysis or Circuit Theory Course taught in Electrical or Computer Engineering Departments. Electric Circuits 9/e is the most widely used introductory circuits textbook of the past 25 years. As this book has evolved over the years to meet the changing learning styles of students, importantly, the underlying teaching approaches and philosophies remain unchanged. | 116 Techniques of Circuit Analysis Example Finding the Thévenin Equivalent of a Circuit with a Dependent Source Find the Thévenin equivalent for the circuit containing dependent sources shown in Fig. . Figure A circuit used to illustrate a Thevenin equivalent when the circuit contains dependent sources. Figure The circuit shown in Fig. with terminals a and b short-circuited. Solution The first step in analyzing the circuit in Fig. is to recognize that the current labeled ix must be zero. Note the absence of a return path for ix to enter the left-hand portion of the circuit. The open-circuit or Thevenin voltage will be the voltage across the 25 ft resistor. With ix 0 VTh ab 20z 25 -500 . The current i is . _ 5 3z _ 5 - 3VTh 1 2000 2000 In writing the equation for z we recognize that the Thevenin voltage is identical to the control voltage. When we combine these two equations we obtain FTh -5 V. To calculate the short-circuit current we place a short circuit across a b. When the terminals a b are shorted together the control voltage v is reduced to zero. Therefore with the short in place the circuit shown in Fig. becomes the one shown in Fig. . With the short circuit shunting the 25 ft resistor all the current from the dependent current source appears in the short so zsc 20 . As the voltage controlling the dependent voltage source has been reduced to zero the current controlling the dependent current source is 5 MOO 2-5 mA- Combining these two equations yields a short-circuit current of zsc 20 50 mA. From zsc and VTh we get V rh 5 q 2 Th X 103 100 ft. zsc -50 Figure illustrates the Thevenin equivalent for the circuit shown in Fig. . Note that the reference polarity marks on the Thevenin voltage source in Fig. agree with the preceding equation for V-fh- 100 il Figure A The Thévenin equivalent for the circuit shown in Fig. . More on Deriving a Thévenin Equivalent 117 assessment problems Objective 5 .

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