Hi All,

Please take a look to the attached circuit.

By the KVL (please correct me if i'm wrong):
Mesh I5
I3R3 + I4R4 + I5R5 = 0
Mesh I4
I8R8 + I2R2 + I3R3 = 30
Mesh I3
I7R7 + I5R5 +I6R6 = -10

I dont know how to do the equation on mesh I1 and I2 because of the current source.
I think I have all the meshe's.

Thanks in advance for your help.

Joaquim

 

You have neglected to take into account the currents flowing in the loops whose current also flows through resistance that are common to a given loop.

I will provide you with the expression for mesh I5 to give you a starting point on how to form the remaining expressions.

mesh I5:

\((I_{5}-I_{4})*R_{3}\ +\ I_{5}*R_{4}\ +\ (I_{5}\ +\ I_{3})*R_{5}\)

hgmjr

 

Thanks hgmjr for the awnser.

I understood what you mean. The influence of the other currents of equal or opposite signal.

But about my primary question . How the solve the mesh with current source? How do I treat the current source?

 

Thanks hgmjr for the awnser.

I understood what you mean. The influence of the other currents of equal or opposite signal.

But about my primary question . How the solve the mesh with current source? How do I treat the current source?

The current loops containing a current source are the easiest of all... they are equal to the current source.

Think about this for a second. The current in a given wire between two nodes must be a constant value, correct? So if I2 - I1 pass through a given wire between two nodes AND a 10 amp current source is in between two nodes... those two values MUST be equal...

For example, in your circuit, 10 = I2 - I1

Now, do KVL for all the loops NOT containing current sources...

But, also make a loop going around both I1 and I2. For example:

4*I1+20+14*I2-14*I4+9*I2-9*I3=0

Does that make sense?

-blazed

 

Thanks Blased

So, if I undestand correctly, the equations for I1 and I2 are 2

I1-I2=10

and the supermesh equation:

4*I1+20+14*I2-14*I4+9*I2-9*I3=0

correct?

Along with this I have the other equations for mesh's I3 I4 and I5

Did I get it?

 

No problem. Everything is correct except the following:

I1-I2=10

If you take another look at your diagram you'll notice that 10 amps is going in the same direction as I2, but in the opposite direction as I1, so 10 = (-I1) + I2 = I2 - I1

Besides that, you've got it.

-blazed (assuming you've corrected your equations using hgmjr's advice...)