## 1 On the Application of Superposition to Dependent Sources in Circuit Analysis

Citations: | 1 - 0 self |

### BibTeX

@MISC{Leach_1on,

author = {W. Marshall Leach},

title = {1 On the Application of Superposition to Dependent Sources in Circuit Analysis},

year = {}

}

### OpenURL

### Abstract

Abstract—Many introductory circuits texts state or imply that superposition of dependent sources cannot be used in linear circuit analysis. Although the use of superposition of only independent sources leads to the correct solution, it does not make use of the full power of superposition. The use of superposition of dependent sources often leads to a simpler solution than other techniques of circuit analysis. A formal proof is presented that superposition of dependent sources is valid provided the controlling variable is not set to zero when the source is deactivated. Several examples are given which illustrate the technique. Index Terms — Circuit analysis, superposition, dependent sources, controlled sources

### Citations

13 |
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- 2005
(Show Context)
Citation Context ...s included at the end of this document. II. INTRODUCTION The author has investigated the presentation of superposition in circuits texts by surveying twenty introductory books on circuit analysis [1]-=-=[20]-=-. Fourteen explicitly state that if a dependent source is present, it is never deactivated and must remain active (unaltered) during the superposition process. The remaining six specifically refer to ... |

12 |
Engineering Circuit Analysis
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(Show Context)
Citation Context ...7) + 3ib = 0 which yields ib = − 7 4 A The total solution is i = ia + ib = 5 4 A This is the same answer obtained by using superposition of the controlled source. B. Example 2 This example comes from =-=[9]-=-. The object is to solve for the voltages v1 and v2 across the current sources in Fig. 2, where the datum node is the lower branch. By superposition, the current i is given by 7 i = 2 7 + 15 + 5 + 3 7... |

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2 |
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(Show Context)
Citation Context ...not using superposition at all leads to a solution with less work. However, the first solution that included the dependent source in the superposition is simpler. C. Example 3 This example comes from =-=[4]-=-. The object is to solve for the current i1 in the circuit of Fig. 3. By superposition, one can write i1 = 30 6 + 4 + 2 Solution for i1 yields Fig. 3. Circuit for example 3. 4 6 + 3 − 8i1 6 + 4 + 2 6 ... |

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2 |
Introduction to Electric Circuits, Sixth Edition
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- 2004
(Show Context)
Citation Context ...w. In each case, it is simpler not to use superposition if the dependent sources remain active. Some of the examples are taken from texts cited in the references. A. Example 1 This example comes from =-=[7]-=-. The object is to solve for the current i in the circuit of Fig. 1. By superposition, one can write i = 24 2 3i 3 − 7 − = 2 − 3 + 2 3 + 2 3 + 2 5 i . ∣ (2) (3)3 Fig. 1. Circuit for example 1. Soluti... |

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1 |
Circuit Anasysis: A Systems Approach, Upper Saddle River, NJ: Pearson/Prentice
- Mersereau, Jackson
- 2006
(Show Context)
Citation Context ...ect is to solve for the voltage vo in the circuit of Fig. 7. By superposition, the voltage v∆ is given by v∆ = −0.4v∆ × 10 + 5 × 10 Fig. 7. Circuit for example 6. G. Example 7 This example comes from =-=[13]-=- in which it is stated in bold, red letters, “Source superposition cannot be used for dependent sources.” The object is to solve for the voltage v as a function of vs and is in the circuit in Fig. 8. ... |