Find the equivalent resistance between A and B. I tried using nodal but there were too many unknown variables.
Please help.
Thank you!
$\begingroup$
$\endgroup$
3
-
2$\begingroup$ This setup is called the Wheatstone bridge. Since the bridge is "balanced" (read he Wikipedia article) there will be no current in the $5 \Omega$ resistor and you can essentially "disconnect" it. $\endgroup$– SivaCommented Oct 3, 2014 at 4:12
-
$\begingroup$ For more information on how to approach "unbalanced" bridges, check here: ibiblio.org/kuphaldt/electricCircuits/DC/DC_10.html $\endgroup$– SivaCommented Oct 3, 2014 at 4:19
-
$\begingroup$ Now I got it..I didn't even think of rearranging it that way. Thank you so much! :) $\endgroup$– ShodaiCommented Oct 3, 2014 at 4:41
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
If you cannot see that this is a Wheatstone bridge (2:4 = 3:6, so the two "arms" are balanced and there is no current in the 5 Ohm resistor) then you can use a simple analysis with just two unknowns: the voltage at the two nodes between A and B, call them x (where 2,4,5 meet) and y (3,5,6 meet). Given an imposed voltage V at A, and 0 (ground) at B, you can now write two equations in three unknowns - namely, the sum of currents flowing out of the nodes x, y must be zero. Now the current flowing out is given by voltage difference between source (the node) and destination (where it's flowing to), so we get:
$$\frac{x-V}{2} + \frac{x-y}{5} + \frac{x}{4} = 0\\ \frac{y-V}{3} + \frac{y-x}{5} + \frac{y}{6} = 0$$
We can pick $V$ to be anything we want it to be - all we care about is the total current that flows because we're looking for the equivalent resistance.
Now $V=I\cdot R$ so $R = \frac{V}{I}$ . We can find $I$ from
$$I = \frac{V-x}{2} + \frac{V-y}{3}$$
If you just set $V=1$ then solving the above equations results in $I=R_{equivalent}$ and you're done.
Leaving that as an exercise... spotting the balanced bridge definitely saves some time.
-
$\begingroup$ I think you have taken x where 2,4,5 meet and y where 3,5,6 meet. But why have you taken the currents as x-V/2 and y-V/3 in the 2 equations instead of V-x/2 and V-y/3? $\endgroup$– ShodaiCommented Oct 3, 2014 at 4:57
-
$\begingroup$ Right on the first point; fixed it. Regarding second point, see if edit makes it clearer. $\endgroup$– FlorisCommented Oct 3, 2014 at 5:00
-
$\begingroup$ Since V is greater than x,y then shouldn't current be flowing from A to x thus on a PD of (V-x)? Same with y, I am getting the current as (V-y)/3 $\endgroup$– ShodaiCommented Oct 3, 2014 at 5:09
-
$\begingroup$ It depends on your definition of "direction of current". If I say "current flows away from x (or y)" then I need to reverse the expression, even though V > x. That's how the sum can be = 0 - I need some of the components to be negative... $\endgroup$– FlorisCommented Oct 3, 2014 at 5:31
-
$\begingroup$ Alright but you can't take all currents to be flowing out, right? (V-x)/2 + (y-x)/5 = x/4 (V-y)/3 - (y-x)/5 = y/6 Adding, we get I = (V-x)/2 + (V-y)/3, same as what you got. $\endgroup$– ShodaiCommented Oct 3, 2014 at 6:14