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Example 4.5 For the circuit in Fig. 4.12, use the superposition theorem to find i. 24 V Solution: ww In this case, we have three sources. Let 42 42 i = i + iz + i3 ww ww where i, iz, and iz are due to the 12-V, 24-V, and 3-A sources respec- tively. To get i, consider the circuit in Fig. 4.13(a). Combining 4 2 (on the right-hand side) in series with 8 2 gives 12 N. The 12 N in parallel with 4 N gives 12 x 4/16 = 3 N. Thus, 12 V O 3 A Figure 4.12 For Example 4.5. 12

Example 4.5 For the circuit in Fig. 4.12, use the superposition theorem to find i. 24 V Solution: ww In this case, we have three sources. Let 42 42 i = i + iz + i3 ww ww where i, iz, and iz are due to the 12-V, 24-V, and 3-A sources respec- tively. To get i, consider the circuit in Fig. 4.13(a). Combining 4 2 (on the right-hand side) in series with 8 2 gives 12 N. The 12 N in parallel with 4 N gives 12 x 4/16 = 3 N. Thus, 12 V O 3 A Figure 4.12 For Example 4.5. 12

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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Example 4.5 For the circuit in Fig. 4.12, use the superposition theorem to find i. 24 V Solution: ww In this case, we have three sources. Let 42 42 i = i + iz + i3 ww ww where i, iz, and iz are due to the 12-V, 24-V, and 3-A sources respec- tively. To get i, consider the circuit in Fig. 4.13(a). Combining 4 2 (on the right-hand side) in series with 8 2 gives 12 N. The 12 N in parallel with 4 N gives 12 x 4/16 = 3 N. Thus, 12 V O 3 A Figure 4.12 For Example 4.5. 12
Transcribed Image Text:Example 4.5 For the circuit in Fig. 4.12, use the superposition theorem to find i. 24 V Solution: ww In this case, we have three sources. Let 42 42 i = i + iz + i3 ww ww where i, iz, and iz are due to the 12-V, 24-V, and 3-A sources respec- tively. To get i, consider the circuit in Fig. 4.13(a). Combining 4 2 (on the right-hand side) in series with 8 2 gives 12 N. The 12 N in parallel with 4 N gives 12 x 4/16 = 3 N. Thus, 12 V O 3 A Figure 4.12 For Example 4.5. 12
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