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3. Replace the circuit 'seen' by the 4 k to find the voltage Vo across the 4 k equivalent instead. resistor by its Thévenin equivalent, and use the equivalent circuit resistor. Repeat the above but replace the circuit 'seen' by its Norton 10 V 2 ΚΩ 14 Solution: VTH = 28 V, RTH = 6 k₁, IN = · 1 ΚΩ 0.5 Vo mA (+) 2 mA ☐ 4 k Vo 3 mA 4.667 mA, RN = 6 ks, V = 11.2 V.

3. Replace the circuit 'seen' by the 4 k to find the voltage Vo across the 4 k equivalent instead. resistor by its Thévenin equivalent, and use the equivalent circuit resistor. Repeat the above but replace the circuit 'seen' by its Norton 10 V 2 ΚΩ 14 Solution: VTH = 28 V, RTH = 6 k₁, IN = · 1 ΚΩ 0.5 Vo mA (+) 2 mA ☐ 4 k Vo 3 mA 4.667 mA, RN = 6 ks, V = 11.2 V.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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3. Replace the circuit 'seen' by the 4 k to find the voltage Vo across the 4 k equivalent instead. resistor by its Thévenin equivalent, and use the equivalent circuit resistor. Repeat the above but replace the circuit 'seen' by its Norton 10 V 1 ΚΩ 0.5 Vo mA 2 ΚΩ 14 Solution: VTH = 28 V, RTH = 6 k₁, IN = · (+) 2 mA ☐ 4 k Vo 3 mA 4.667 mA, RN = 6 ks, V = 11.2 V.
Transcribed Image Text:3. Replace the circuit 'seen' by the 4 k to find the voltage Vo across the 4 k equivalent instead. resistor by its Thévenin equivalent, and use the equivalent circuit resistor. Repeat the above but replace the circuit 'seen' by its Norton 10 V 1 ΚΩ 0.5 Vo mA 2 ΚΩ 14 Solution: VTH = 28 V, RTH = 6 k₁, IN = · (+) 2 mA ☐ 4 k Vo 3 mA 4.667 mA, RN = 6 ks, V = 11.2 V.
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