Mesh Analysis Examples

Example A 40V 6.364A 20V 1.818A 4 6 I1I1 I2I2

Using Mathcad

Using PSpice I 1 = A I 2 = A

Example 2: By Inspection 8.863A 12V 6.549A A V 18V V 2 I2I2 I1I1 I3I3

Using Mathcad: By Inspection

Using PSpice

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

Using Mathcad Mesh 1:

Using PSpice I 1 = 250  A

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:

Using Mathcad

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

Example 5: Supermesh

2.5I b 20V 2I a A IaIa IbIb I1I1 I2I2 I3I3 I 1 = 5.431A I 2 = A I 3 = -1.81A Ans:

Using Mathcad: equations

Using Mathcad: Results

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

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 + -

Using Mathcad

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

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 = V

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

Using Mathcad: Equations

Using Mathcad: Results

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

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

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

Power absorbed by passive elements x 2.5I b 20V 2I a A IaIa IbIb I1I1 I2I2 I3I VcVc + - V 5A The power absorbed by the resistor is always positive; we can always use P = I 2 R

Power absorbed by all elements