1. Mesh Analysis Examples

2. Example 1 26.364A 40V 6.364A 20V 1.818A 4 6 I1I1 I2I2

3. Using Mathcad

4. Using PSpice I 1 = -1818 A I 2 = -6.364 A

5. Example 2: By Inspection 8.863A 12V 6.549A 4 6.235A 1 1 24V 18V 30 1 2 6V 2 I2I2 I1I1 I3I3

6. Using Mathcad: By Inspection

7. Using PSpice

8. Example 3 4k 2mA 3V 4k 6k 2k 4mA -2mA 4mA I1I1

9. Using Mathcad Mesh 1:

10. Using PSpice I 1 = 250  A

11. Example 4: Supermesh 3 20V 3 6A 4 6 I2I2 I1I1 Mesh 1 and mesh 2 form supermesh I 1 = -1.6 A I 2 = 4.4 A Ans:

12. Using Mathcad

13. Using PSpice I 1 = -1.6 AI 2 = 4.4 A Ans:

14. Example 5: Supermesh

15. 2.5I b 20V 2I a + - 8 80 40 5A IaIa IbIb I1I1 I2I2 I3I3 I 1 = 5.431A I 2 = -3.621A I 3 = -1.81A Ans:

16. Using Mathcad: equations

17. Using Mathcad: Results

18. Using PSpice I 1 = 5.431A I 2 = -3.621A I 3 = -1.81A Ans:

19. Find V a and I x Problem 1 i1i1 3A 6V - 22 22 44 88 11 i2i2 i3i3 + VaVa IxIx + -

20. Using Mathcad

21. I x = -I 1 = -3.474 A To Find I x i1i1 3A 6V - 22 22 44 88 11 i2i2 i3i3 + VaVa IxIx + -

22. Mesh 1: -6+(I 1 -I 3 )  2+(1  3)+V a =0 To Find V a i1i1 3A 6V - 22 22 44 88 11 i2i2 i3i3 + VaVa IxIx + - + - + - + - V8V8 V4V4 V1V1 V2V2 - + Mesh 2: -(1  3)+ (I 2 -I 3 )4+8I 2 -V a =0 or V a = -1.737V

23. Problem 2 Find V xy, V c, and the power absorbed by all elements. x 2.5I b 20V 2I a + - 8 80 40 5A IaIa IbIb I1I1 I2I2 I3I3 10 y + - VcVc

24. Using Mathcad: Equations

25. Using Mathcad: Results

26. To Find V xy V xy, V c x 2.5I b 20V 2I a + - 80 5A IbIb I1I1 I2I2 I3I3 10 y + - VcVc + - -V xy -2I a +10I 3 = 0 V xy V xy = -2I a +10I 3

27. To Find V c x 2.5I b 20V 2I a + - 80 5A IbIb I1I1 I2I2 I3I3 10 y + - VcVc + - +V c +(I 2 -I 3 )40-20= 0 V 40 V c = -(I 2 -I 3 )40+20 40

28. Power absorbed by active elements x 2.5I b 20V 2I a + - 8 80 40 5A IaIa IbIb I1I1 I2I2 I3I3 10 y + - VcVc + - V 5A For the active elements, we have to follow PSC to calculate the power absorbed.

29. Power absorbed by passive elements x 2.5I b 20V 2I a + - 8 80 40 5A IaIa IbIb I1I1 I2I2 I3I3 10 + - VcVc + - V 5A The power absorbed by the resistor is always positive; we can always use P = I 2 R

30. Power absorbed by all elements