What is the equivalent resistance across points a and b?

[Piyu]

1. find the equivalent resistance across points a and b

The Attempt at a Solution

I've used kirchhoff junction and loop rule to get a whole chunk of simultaneous equations but i can't seem to lay the finisher.

Assigning I_1 to I_4 in the clock wise manner starting from the 5 ohms resistor, and I_5 to the middle 2 ohms resistor. V = potential difference between a and b.

V=5I₅+4I₂=3I₄+3I₅
I=I₁+I₄=I₂+I₅

Ill end up with R=V/I = 15V/(8V-15I₅-12I₂. and here I am pretty much stuck with no way to change the remaining 2 to equations of V

 

[Bhumble]

Parallel resistors are inversely proportional. Series resistors are directly proportional.

Parallel
\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}

Series
R_{total} = R_1 + R_2

Redraw the diagram to make it easier to distinguish between series and parallel.

 

[Piyu]

I don't think we can approach it from using the resistor sums for series and parallel because i can't seem to redraw it in any way that i can use the 2 forumulas. Maybe, too much staring at it has made me cock-eyed :P

 

[Bhumble]

You're right this is an annoying circuit. It is a wheatstone bridge. But you can use Thevenize it to make it simpler. Should end up as
R_{total} = \frac{1}{\frac{1}{3 \Omega} + \frac{1}{3 \Omega}} + \frac{1}{\frac{1}{5 \Omega} + \frac{1}{4 \Omega}} + 2 \Omega

Just google wheatstone bridge and you should find a decent explanation.

 

[gneill]

Obviously there are several methods available for approaching this problem. One might, for example, place a 1V voltage source across a-b and write the Kirchoff loop equations for the three loops, solving for the current through the source voltage. V/I gives your resistance.

Another approach is to transform one of the ∆ shaped configurations of resistors to a Y sharped one, thus allowing one to proceed with the usual parallel and serial simplifications down to a single resistor.

Here's a http://www.allaboutcircuits.com/vol_1/chpt_10/13.html" .