Kirchoff's Rule(How do I find equivalent resistance?)

[jawhnay]

Homework Statement

Please take a look at the attachment to see the circuit. My question is how do I find the equivalent resistance of this circuit? I have done the other two parts but this part has got me stumped. I also drew my currents on the picture.

Homework Equations

The Attempt at a Solution

I attempted to categorize which resistors were in parallel and in series to each other but that just got me really confused... If anyone can walk me through this problem it would be greatly appreciated.

Here are my loop equations:

-9I₁+3I₂+9I₃+8I₄ = -12
-3I₃ = -12
-6I₁-4I₂+6I₃+6I₄ = -12
-I₁+I₂+I₃-5I₄ = -12

I₁ = 7.324841
I₂ = 0.878981
I₃ = 4
I₄ = 1.910828

current across the 5 ohm resistor = 1.911 amps
V_ab = -8.484 volts

 

[phinds]

Do you know how to write loop equations? You need to show some effort on your own so we know where you are stumped. Just giving the answer is against forum rules.

EDIT: "walking you through it" is equivalent to giving the answer. We're more into hints.

 

[jawhnay]

yes, I know how to write loop equations. However, I thought the equivalent resistance was just adding up the resistances so I didn't think I would need to show my loop equations. I have edited my first post with the loop equations and edited picture.

 

[phinds]

Since you have the current being drawn out of the 12V supply, you have the equivalent resistance, yes?

The problem is that "adding up the resistances" just doesn't work unless everything is simple series/parallel combinations and you've already figured out that that is not the case with this circuit else you would have been able to do it.

 

[jawhnay]

oh my gosh, I cannot believe I didn't see it. Thank you for the help!

 

[gneill]

Since you have the current being drawn out of the 12V supply, you have the equivalent resistance, yes?

The problem is that "adding up the resistances" just doesn't work unless everything is simple series/parallel combinations and you've already figured out that that is not the case with this circuit else you would have been able to do it.

Actually, if you note that there are "missing" resistors on the diagonals you might see that direct simplification is quite possible. Easy in fact