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Step 2: After Step 1 we examine the problem till we are confident with a plan for finding the unknown. I plan to find everything by three laws. It may not be necessary to find everything, if we consider the only unknown is V, but for illustration, I shall find everything anyway. . Вy KCL: 3 At Node 3 i1-İ2 +1 A 10 Ω 1 A + At Node O i3-İ2 + 1 A is It is obvious that i3- iş = i1 (Is it to you?) 2 202 4 By KVL: Take a loop from 3 back to 3 via: 3-4-2-1-3: Vz+V3-V,+V1= 0 Another loop from 3 to 4 via V, back to 3 via V2: V4-V2 = 0 You may want to take another loop as the first but from 3 to 4 via V4. Try it and see that it can be derived from the two equations Next we write the conditions on these unknowns using the laws. By Ohm's law: V1= i,x 10; V2 = izx 2; V3 = i3x 20; V4= (1 A) x 10 ohms %3D above.

Step 2: After Step 1 we examine the problem till we are confident with a plan for finding the unknown. I plan to find everything by three laws. It may not be necessary to find everything, if we consider the only unknown is V, but for illustration, I shall find everything anyway. . Вy KCL: 3 At Node 3 i1-İ2 +1 A 10 Ω 1 A + At Node O i3-İ2 + 1 A is It is obvious that i3- iş = i1 (Is it to you?) 2 202 4 By KVL: Take a loop from 3 back to 3 via: 3-4-2-1-3: Vz+V3-V,+V1= 0 Another loop from 3 to 4 via V, back to 3 via V2: V4-V2 = 0 You may want to take another loop as the first but from 3 to 4 via V4. Try it and see that it can be derived from the two equations Next we write the conditions on these unknowns using the laws. By Ohm's law: V1= i,x 10; V2 = izx 2; V3 = i3x 20; V4= (1 A) x 10 ohms %3D above.

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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Step 2: After Step 1 we examine the problem till we are confident with a plan for finding the unknown. I plan to find everything by three laws. It may not be necessary to find everything, if we consider the only unknown is V, but for illustration, I shall find everything anyway. By KCL: At Node (3 + V1 3 i1-iz +1 A 10 Ω 1 A + At Node (4 i3 = İ2 + 1 A 20|i. V2 iş It is obvious that iz-iş = i1 (Is it to you?) V3 + 2 202 4 By KVL: Take a loop from 3 back to 3 via: 3-4-2-1-3: Vz+V3-V,+V1= 0 Another loop from 3 to 4 via V, back to 3 via V2: V4- V2 = 0 You may want to take another loop as the first but from 3 to 4 via V4. Try it and see that it can be derived from the two equations 13 %3D Next we write the conditions on these unknowns using the laws. By Ohm's law: V1 = i,x 10; V2= i2x 2; V3 = i3x 20; V4 = (1 A) x 10 ohms above.
Transcribed Image Text:Step 2: After Step 1 we examine the problem till we are confident with a plan for finding the unknown. I plan to find everything by three laws. It may not be necessary to find everything, if we consider the only unknown is V, but for illustration, I shall find everything anyway. By KCL: At Node (3 + V1 3 i1-iz +1 A 10 Ω 1 A + At Node (4 i3 = İ2 + 1 A 20|i. V2 iş It is obvious that iz-iş = i1 (Is it to you?) V3 + 2 202 4 By KVL: Take a loop from 3 back to 3 via: 3-4-2-1-3: Vz+V3-V,+V1= 0 Another loop from 3 to 4 via V, back to 3 via V2: V4- V2 = 0 You may want to take another loop as the first but from 3 to 4 via V4. Try it and see that it can be derived from the two equations 13 %3D Next we write the conditions on these unknowns using the laws. By Ohm's law: V1 = i,x 10; V2= i2x 2; V3 = i3x 20; V4 = (1 A) x 10 ohms above.
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