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SuperEx

# SuperEx - Superposition Examples The following examples...

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Superposition Examples The following examples illustrate the proper use of superposition of dependent sources. All superposition equations are written by inspection using voltage division, current division, series-parallel combinations, and Ohm’s law. In each case, it is simpler not to use superposition if the dependent sources remain active. Example 1 Theobjectistoso lveforthecurrent i in the circuit of Fig. 1. By superposition, one can write i = 24 3+2 7 2 3+2 3 i 3+2 =2 3 5 i Solution for i yields i = 2 1+3 / 5 = 5 4 A Figure 1: Circuit for example 1. If superposition of the controlled source is not used, two solutions must be found. Let i = i a + i b ,where i a is the current with the 7A source zeroed and i b is the current with the 24 V source zeroed. By superposition, we can write i a = 24 3+2 3 i a 3+2 i b = 7 2 3+2 3 i b 3+2 Solution for i a and i b yields i a = 24 3+2 1+ 3 3+2 =3A i b = 7 2 3+2 1+ 3 3+2 = 7 4 A The solution for i is thus i = i a + i b = 5 4 A This is the same answer obtained by using superposition of the controlled source. Example 2 The object is to solve for the voltages v 1 and v 2 across the current sources in Fig. 2, where the datum node is the lower branch. By superposition, the current i is given by i =2 7 7+15+5 + 3 7+15+5 +4 i 7+15 7+15+5 = 17 27 + 88 27 i Solution for i yields i = 17 / 27 1 88 / 27 = 17 61 A Although superposition can be used to solve for v 1 and v 2 , it is simpler to write v 2 =5 i = 1 . 393 V v 1 = v 2 (4 i i )15=11 . 148 V

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Figure 2: Circuit for example 2. Example 3 Theobjectistoso lveforthecurrent i 1 in the circuit of Fig. 3. By superposition, one can write i 1 = 30 6+4+2 +3 4 6+4+2 8 i 1 6 6+4+2 = 42 12 4 i 1 Solution for i 1 yields i 1 = 42 / 12 1+4 =0 . 7A Figure 3: Circuit for example 3.
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SuperEx - Superposition Examples The following examples...

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