352 i1 2i2 1 2 1 1 a thus i1 i1 1 a method

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Unformatted text preview: d the branch currents I1 , I2 , and I3 using mesh analysis. 6Ω I3 Solution: We first obtain the mesh currents using KVL. For mesh 1, 10 Ω 15 V + − i1 −15 + 5i1 + 10(i1 − i2 ) + 10 = 0 4Ω i2 or + 10 V − 3i1 − 2i2 = 1 (3.5.1) For mesh 2, Figure 3.18 6i2 + 4i2 + 10(i2 − i1 ) − 10 = 0 For Example 3.5. or i1 = 2i2 − 1 (3.5.2) M E T H O D 1 Using the substitution method, we substitute Eq. (3.5.2) into Eq. (3.5.1), and write 6i2 − 3 − 2i2 = 1 ⇒ i2 = 1 A From Eq. (3.5.2), i1 = 2i2 − 1 = 2 − 1 = 1 A. Thus, I1 = i1 = 1 A, METHOD 2 I2 = i2 = 1 A, I3 = i1 − i2 = 0 To use Cramer’s rule, we cast Eqs. (3.5.1) and (3.5.2) in matrix form as 3 −1 −2 2 i1 1 = i2 1 We obtain the determinants = 1 = 1 1 3 −1 −2 =6−2=4 2 −2 = 2 + 2 = 4, 2 2 = 3 −1 1 =3+1=4 1 Thus, i1 = 1 = 1 A, i2 = 2 =1A as before. PRACTICE PROBLEM 3.5 2Ω i1 12 V + − Figure 3.19 Calculate the mesh currents i1 and i2 in the circuit of Fig. 3.19. Answer: i1 = 2 A, i2 = 0 A. 3 + − 8V 3Ω...
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This note was uploaded on 07/16/2012 for the course KA KA 2000 taught by Professor Bkav during the Spring '12 term at Cambridge.

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