# Kirchhoff analysis - Help needed

Discussion in 'Homework Help' started by jboavida, Nov 22, 2008.

1. ### jboavida Thread Starter Member

Jul 10, 2008
23
0
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.

Joaquim

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2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
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

3. ### jboavida Thread Starter Member

Jul 10, 2008
23
0
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?

Last edited: Nov 22, 2008
4. ### blazedaces Active Member

Jul 24, 2008
130
0
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

5. ### jboavida Thread Starter Member

Jul 10, 2008
23
0
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?

6. ### blazedaces Active Member

Jul 24, 2008
130
0
No problem. Everything is correct except the following:

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.