EE201 Lecture 9 P 1 1A Example i2 2 10V 1A 4 3A i1 12V I i 3 1 i5 i6 7V 3 i4 2A Find I for this circuit 5V 2A EE201 Lecture 9 P 2 Solution Four mesh currents that can be written directly i1 1 A i2 1A i3 2 A i4 2A There is a constraint on i5 and i6 due to source i 6 i5 3 A To find i5 and i6 define supermesh around 5 and 6 KVL supermesh 12 4 i5 i1 2 i5 i2 10 1 i6 i3 5 7 3 i6 i4 0 Plugging in mesh currents reduces this to 6 i5 4 i6 12 i5 2 4 A I i3 i6 2 0 6 i6 0 6 A I 2 6 A EE201 Lecture 9 P 3 Mesh Analysis with Dependent Current Source R 15A i1 1 3 Vx 1 9 Vx i2 i3 2 Mesh equations KVL 1 i1 15A KVL 2 1 i2 i1 2i2 3 i2 i3 0 I source 1 9 Vx i3 i1 Vx 3 i3 i2 i3 i2 3 i3 i1 2 1 EE201 Lecture 9 P 4 Simultaneous equations to solve i1 15 3i1 i2 2i3 0 i1 6i2 3i3 0 i2 11 A and i3 17 A Example Find mesh currents i1 and i2 4 23 V 0 9 i2 i1 1 8V EE201 Lecture 9 P 5 Solution KVL1 KVL2 23 4i1 1 i1 i2 0 8 0 9i2 1 i2 i1 0 Solve these equations simultaneously i1 2 A i2 4 2 A Example Find mesh currents i2 and i3 2 V i2 x 1 4 5A 3 0 25 Vx i1 2 i3 1 EE201 Lecture 9 P 6 Solution Mesh current i1 can be written directly i1 5 A KVL of mesh 2 KVL2 1 i2 i1 2i2 3 i2 i3 0 6 i2 3i3 5 There is a constraint on i1 and i3 from dependent source i3 i1 0 25 Vx 0 25 2i2 0 5i2 Solving simultaneous equation i2 8 3 A i3 11 3 A
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