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

 

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

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.

 

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????

 

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.

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.

 

or reduced likelihood of making an error when writing the mesh equations.

Thanks,
That's probably why they taught us using one direction.

 

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

 

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.

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?

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

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.

 

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?

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?)

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.

Ok, got it. thanks

 

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?

Nop don't know that.

If the voltage sources weren't present, there would be no currents at all; do you understand why?

 

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