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1. Apply KCL to find the value of I in Fig. 2.36. [8 A] +15 V- |C } Tiov 5A 10 V A B 3A Fig. 2.36 Fig. 2.37 2. Applying Kirchhoff's voltage law, find V, and V, in Fig. 2.37. 3. Find the values of currents I, and I, in the network of Fig. 2.38. [V½ = 10 V; V,=5 V] [I, = 4 A ; I, = 5 A] 9V 18 V /4 4 = 6A 12 V 10 V 2A 9A 3V 3 Fig. 2.38 Fig. 2.39 4. Use Kirchhoff's law, to find the values of voltages V, and V, in the network shown in Fig. 2.39. [V = 2 V ; V, = 5 V]

1. Apply KCL to find the value of I in Fig. 2.36. [8 A] +15 V- |C } Tiov 5A 10 V A B 3A Fig. 2.36 Fig. 2.37 2. Applying Kirchhoff's voltage law, find V, and V, in Fig. 2.37. 3. Find the values of currents I, and I, in the network of Fig. 2.38. [V½ = 10 V; V,=5 V] [I, = 4 A ; I, = 5 A] 9V 18 V /4 4 = 6A 12 V 10 V 2A 9A 3V 3 Fig. 2.38 Fig. 2.39 4. Use Kirchhoff's law, to find the values of voltages V, and V, in the network shown in Fig. 2.39. [V = 2 V ; V, = 5 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
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1. Apply KCL to find the value of I in Fig. 2.36. [8 A] +15 V- |C } Tiov 5A 10 V A B 3A Fig. 2.36 Fig. 2.37 2. Applying Kirchhoff's voltage law, find V, and V, in Fig. 2.37. 3. Find the values of currents I, and I, in the network of Fig. 2.38. [V½ = 10 V; V,=5 V] [I, = 4 A ; I, = 5 A] 9V 18 V /4 4 = 6A 12 V 10 V 2A 9A 3V 3 Fig. 2.38 Fig. 2.39 4. Use Kirchhoff's law, to find the values of voltages V, and V, in the network shown in Fig. 2.39. [V = 2 V ; V, = 5 V]
Transcribed Image Text:1. Apply KCL to find the value of I in Fig. 2.36. [8 A] +15 V- |C } Tiov 5A 10 V A B 3A Fig. 2.36 Fig. 2.37 2. Applying Kirchhoff's voltage law, find V, and V, in Fig. 2.37. 3. Find the values of currents I, and I, in the network of Fig. 2.38. [V½ = 10 V; V,=5 V] [I, = 4 A ; I, = 5 A] 9V 18 V /4 4 = 6A 12 V 10 V 2A 9A 3V 3 Fig. 2.38 Fig. 2.39 4. Use Kirchhoff's law, to find the values of voltages V, and V, in the network shown in Fig. 2.39. [V = 2 V ; V, = 5 V]
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