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1.15 Through repeated application of Thévenin's theorem, find the Thévenin equivalent of the circuit in Fig. Pl.15 between node 4 and ground, and hence find the current that flows through a load resistance of 3 k2 connected between node 4 and ground.

1.15 Through repeated application of Thévenin's theorem, find the Thévenin equivalent of the circuit in Fig. Pl.15 between node 4 and ground, and hence find the current that flows through a load resistance of 3 k2 connected between node 4 and ground.

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
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1.15 Through repeated application of Thévenin's theorem, find the Thévenin equivalent of the circuit in Fig. Pl.15 between node 4 and ground, and hence find the current that flows through a load resistance of 3 k2 connected between node 4 and ground. 1 20 k2 2 20 k2 3 20 k2 4 10 V 20 k2 20 ΚΩ 20 kΩ Figure P1.15
Transcribed Image Text:1.15 Through repeated application of Thévenin's theorem, find the Thévenin equivalent of the circuit in Fig. Pl.15 between node 4 and ground, and hence find the current that flows through a load resistance of 3 k2 connected between node 4 and ground. 1 20 k2 2 20 k2 3 20 k2 4 10 V 20 k2 20 ΚΩ 20 kΩ Figure P1.15
20 kN 20 kN 20 kN 4 10 V 20 k2 20 k2 20 k2 12.31 kN 10 kN 2 20 k2 20 kN 3 4 30 kΩ 0.77 V 3 kN 20 kN 20 k2 20 10 X 20 + 20 = 5 V Now, when a resistance of 3 k2 is connected (20//30) = 12 klN 3 -어 20 k2 4 between node 4 and ground, 0.77 I = 12.31 +3 = 0.05 mA 32 k2 20 20 kΩ 5X 20 + 30 = 2V Thévenin equivalent: (20//32) = 12.31 kN 20 2 X 20 +32 = 0.77 V (cont d on the next page) PIC•COLLAGE
Transcribed Image Text:20 kN 20 kN 20 kN 4 10 V 20 k2 20 k2 20 k2 12.31 kN 10 kN 2 20 k2 20 kN 3 4 30 kΩ 0.77 V 3 kN 20 kN 20 k2 20 10 X 20 + 20 = 5 V Now, when a resistance of 3 k2 is connected (20//30) = 12 klN 3 -어 20 k2 4 between node 4 and ground, 0.77 I = 12.31 +3 = 0.05 mA 32 k2 20 20 kΩ 5X 20 + 30 = 2V Thévenin equivalent: (20//32) = 12.31 kN 20 2 X 20 +32 = 0.77 V (cont d on the next page) PIC•COLLAGE
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