# analyzing a circuit with kirchhoff's laws

Discussion in 'Homework Help' started by drk, Nov 13, 2009.

1. ### drk Thread Starter Active Member

Mar 8, 2008
41
1
Hello,
there's this exercise that i have to resolve using kirchhoffs laws, but I'm not sure I'm doing it right(the schematic is attached).
What i did until now was:
I saw that this circuit has 4 nodes, and so i can do 3 KCL.
So I got:
i4+i2-i3=0
i3+i5-i1-i7=0
i7+i6-i8=0

i have 8 unknown currents, so i have to have 8 equations, i need 5 KVL equations. this is where I'm not sure i got it right.

what i did:
e1+e2+e3+10=0
e3-e5+e4=0
e5+4.5+i8=0
e2-e4-i8-e6=0
10+e1+e2-e4-i8-4.5=0

R1=330 ohm
R2=470 ohm
R3=330 ohm
R4=680 ohm
R5=5600 ohm
R6=2200 ohm

so it comes:
330*i1+470*i2+330*i3=-10
330*i3-5600*i5+680*i4=0
5600*i5+i8=-4.5
470*i2-680*i4-i8-2200*i6=0
330*i1+470*i2-680*i4-i8=-5.5

Now, i can't resolve this because my calculator only goes until 6 unknowns, and this has 8.
can anyone check this out?
thanks

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2. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,724
496
Where are the voltage sources in these equations?

Regarding the second part of your solution, why don't you use the mesh current method instead of the branch current method? See: http://www.allaboutcircuits.com/vol_1/chpt_10/index.html

You'll be using KCL but you'll only have to deal with 4 unknowns instead of 8.

Last edited: Nov 14, 2009
3. ### hobbyist AAC Fanatic!

Aug 10, 2008
887
88
This might be irrelevent, but I think that when doing mesh analysis the assumed current direction is used in the same direction in all the loops.

Correct me if I'm wrong, but It looks like you went clockwise in first loop, then counter clockwise second loop,

Then the third loop your including current alone, ect...

I was taught in the books that the mesh equations should be done like this for KVL.

If you assume clockwise direction for each loop.

E6 + E4 - E2 = 0
E1 + E2 + E3 + 10 = 0
-E3 - E4 + E5 = 0
-E5 - 4.5 = 0
E1 + E6 - 4.5 + 10 = 0

Does that look right?. anyone????

Last edited: Nov 13, 2009
4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
You can choose whatever direction you wish for any particular loop - it can be a mixture of clockwise or anti-clockwise. Most people probably choose one direction for convenience &/or reduced likelihood of making an error when writing the mesh equations.

5. ### hobbyist AAC Fanatic!

Aug 10, 2008
887
88
Thanks,
That's probably why they taught us using one direction.

6. ### drk Thread Starter Active Member

Mar 8, 2008
41
1
The Electrician:
In those equations i simply used the rule, that the currents entering a node equals the currents exiting. don't know if its right though.

hmm so if asked to do a circuit with kirchhoffs laws, i can choose between the mesh and branch method?

hobbyist:
the directions were already like that in the exercise. i also prefer to do it all the same direction.

i'll try to do it with the mesh current.
thanks

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,724
496
You do understand that the voltage sources will have an effect on currents in the network, don't you? Since that is the case, then a solution to the network will have to involve the voltage sources, right?

The branch, nodal, mesh and loop methods all use one or the other of Kirchoff's laws, so if your instructor just said to use Kirchoff's laws, then I would guess that you could use any one of the four methods.

8. ### drk Thread Starter Active Member

Mar 8, 2008
41
1
Nop don't know that. You're saying that when i do the KVL equations, these have to be related to the KCL done before(for example, if I choose Ir1 on KCL, i have to include Vr1?)

Ok, got it. thanks

9. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,724
496
If the voltage sources weren't present, there would be no currents at all; do you understand why?

10. ### drk Thread Starter Active Member

Mar 8, 2008
41
1
ahh yes, you're saying that if there isn't voltage, there's no current.
ok i got it.