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## Homework Statement

Find i0 in http://images3a.snapfish.com/232323232fp73395>nu=52::>379>256>WSNRCG=3359;24399347nu0mrj" using mesh current analysis.

Note that the ^i1, ^i2 and ^i3 are to label my currents and show they are flowing clockwise. i0--> is the direction of the i0 current. The current source is 3A.

## Homework Equations

KVL

KCL

Ohm's law

## The Attempt at a Solution

So i know that the current sources makes a supermesh and that you basically remove it from the circuit and do mesh analysis 'around the path of the supermesh' and you will get a constraint equation from the supermesh for your 3rd equation.

So for i1 I got:

[tex] 2i_{1} + 6 + 4(i_{1}-i_{3})+ (i_{1}-i_{2}) = 0 [/tex]

[tex] 2i_{1} + 4i_{1} - 4i_{3} + i_{1} - i_{2} = -6 [/tex]

[tex] 8i_{1} - i_{2}-4i_{3}=-6 [/tex]

So the supermesh i got:

[tex] 5i_{2} + (i_{2}-i_{1}) + 4(i_{3}-i_{1})+12=0[/tex]

[tex] 5i_{2} + i_{2} - i_{1} + 4i_{3}-4i_{1}=-12 [/tex]

[tex] -5i_{1} + 6i_{2} + 4i_{3} = -12 [/tex]

and the constraint equation:

[tex] i_{3}=i_{2}+3 [/tex]

[tex] 0i_{1}-i_{2}+i_{3} = 3 [/tex]

which makes a 3x4 matrix now i got:

[tex] i_{1}=-1.09a [/tex]

[tex] i_{2}=-2.945a [/tex]

[tex] i_{3}=.054a [/tex]

and i believe that i0 in the circuit should be i2-i1 but they are looking for -1.73a for i0 which is close to but not what you get with my numbers so where did I go wrong?

thanks!

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