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Example Use the superposition theorem to find v in the circuit of Fig. 4.6. 8Ω Solution: Since there are two sources, let 6 V 4Ω 43 A v = U1 + 2

Example Use the superposition theorem to find v in the circuit of Fig. 4.6. 8Ω Solution: Since there are two sources, let 6 V 4Ω 43 A v = U1 + 2

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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Superposition: Hi, can you find "v" using superposition with mesh analysis only? No current and voltage division please.
Keep in mind that superposition is based 8R reason, it is not applicable to the effect on power due to each source, because the power absorbed by a resistor depends on the squ'are of the voltage or current. If the power value is needed, the current through (or voltage across) the element must be calculated first using superposition. Example 4.3 Use the superposition theorem to find v in the circuit of Fig. 4.6. 82 Solution: Since there are two sources, let 6 V 4Ω 3 A v = U, + V2 where v and vz are the contributions due to the 6-V voltage syurce and the 3-A current source, respectively. To obtain v,, we set the current source to zero, as shown in Fig. 4.7(a). Applying KVL to the loop in Fig. 4.7(a) gives Figure 4.6 For Example 4.3. 12i, - 6 = 0 i = 0.5 A 132 Chapter 4 Circuit Theorems 89 TI
Transcribed Image Text:Keep in mind that superposition is based 8R reason, it is not applicable to the effect on power due to each source, because the power absorbed by a resistor depends on the squ'are of the voltage or current. If the power value is needed, the current through (or voltage across) the element must be calculated first using superposition. Example 4.3 Use the superposition theorem to find v in the circuit of Fig. 4.6. 82 Solution: Since there are two sources, let 6 V 4Ω 3 A v = U, + V2 where v and vz are the contributions due to the 6-V voltage syurce and the 3-A current source, respectively. To obtain v,, we set the current source to zero, as shown in Fig. 4.7(a). Applying KVL to the loop in Fig. 4.7(a) gives Figure 4.6 For Example 4.3. 12i, - 6 = 0 i = 0.5 A 132 Chapter 4 Circuit Theorems 89 TI
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