Solving a Circuit Diagram with KVL & KCL

[shayaan_musta]

Here is my circuit diagram.
http://s1227.photobucket.com/albums...n=view&current=untitled-1.jpg&t=1299237316875
I am unable to solve my circuit with the help of Kirchhoff's law:
Here is my KVL equations:
24-v₁-v₃-v₈=0------(1)
v₃-v₂-v₄-v₇=0------(2)
v₄-v₅-v₆=0------(3)
24-v₁-v₂-v₅-v₆=0------(outside loop i.e. 4)

And this is my KCL equations:
I_s=I₂+I₃------(5)
I₂=I₄+I₅------(6)
I₅=I₆------(7)
I₄=I₇------(8)
I₈=I₃+I₇-------(9)
I_s=I₈+I₆-------(10)



I have another thing to ask you experts that in circuits I have show two paths for I₆ one is blue and other is red. I want to ask that form which way I₆ will pass? either by blue or red??
My KCL equations are from the way i.e. blue.

Are my KCL & KVL equations are correct or not?
Kindly help me.

 

[berkeman]

Here is my circuit diagram.
http://s1227.photobucket.com/albums...n=view&current=untitled-1.jpg&t=1299237316875
I am unable to solve my circuit with the help of Kirchhoff's law:
Here is my KVL equations:
24-v₁-v₃-v₈=0------(1)
v₃-v₂-v₄-v₇=0------(2)
v₄-v₅-v₆=0------(3)
24-v₁-v₂-v₅-v₆=0------(outside loop i.e. 4)

And this is my KCL equations:
I_s=I₂+I₃------(5)
I₂=I₄+I₅------(6)
I₅=I₆------(7)
I₄=I₇------(8)
I₈=I₃+I₇-------(9)
I_s=I₈+I₆-------(10)



I have another thing to ask you experts that in circuits I have show two paths for I₆ one is blue and other is red. I want to ask that form which way I₆ will pass? either by blue or red??
My KCL equations are from the way i.e. blue.

Are my KCL & KVL equations are correct or not?
Kindly help me.

It looks like R7 and R8 are short-circuited across the bottom -- is that correct? That certainly would simplify the problem...

 

[shayaan_musta]

It looks like R7 and R8 are short-circuited across the bottom -- is that correct? That certainly would simplify the problem...

no no it is not short circuited.
it is connected with R₃ and making a junction. I have corrected the mistake:
http://s1227.photobucket.com/albums...n=view&current=untitled-1.jpg&t=1299323931498

Now kindly tell me sir.

 

[berkeman]

no no it is not short circuited.
it is connected with R₃ and making a junction. I have corrected the mistake:
http://s1227.photobucket.com/albums...n=view&current=untitled-1.jpg&t=1299323931498

Now kindly tell me sir.

The blue line labelled I6 is still shorting the two resistors. I'm not understanding the diagram...

 

[vk6kro]

There is a lot of simplification available in this circuit.

R7 and R8 are in parallel and in series with R3.

R4 is in parallel with (R5 + R6).

You can probably just use Ohm's Law from there.