I was looking through some old electronics books of mine, and I can't
seem to figure out the equation to calculate the following circuit:
o



 
R R
1 4
 
R 3
 
R R
2 5
 



o
I just can't figure out how R3 fits in. I would guess that it is
somehow in parallel, but I'm not sure with what. Could someone write
out the equation for solving the total resistance of this circuit? I'm
very confused.
Thanks a bunch,
Danny H.
Assume:
o Assume +V here

 V(1)
,,
 
R R
1 4
 
V(2) +R 3+ V(3)
 
R R
2 5
 
''
 V(4)

o Assume V=0 here
If the ratios of R1/R2 and R4/R5 are equal, it becomes trivial because
no current flows through R3 and you can eliminate it from your
analysis.
But don't be too hard on yourself if they aren't equal, as this is the
everpopular unbalanced Wheatstone bridge problem and it takes an
amount of algebra to solve it via simultaneous equations and a branch
current analysis. (Or you can use mesh analysis, which is perhaps a
little easier, algebraically speaking.)
Here is how a spice program might solve the problem. First, spice
uses conductances for the resistors, instead of resistance values.
(Which is a reason why spice hates resistances of 0 ohms.)
So it would do something like:
Treat the resistances as conductances,
G1 = 1/R1
G2 = 1/R2
G3 = 1/R3
G4 = 1/R4
G5 = 1/R5
Set known voltages,
V(1) = V
V(4) = 0 (Spice needs a ground ref, so it might as well be here.)
Set up an equation for the rest (KCL/KVL):
V(2)*(G1+G2+G3)V(3)*G3V(1)*G1V(4)*G2 = 0
V(3)*(G3+G4+G5)V(2)*G3V(1)*G4V(4)*G5 = 0
One way of viewing the above setup for these two "= 0" equations (and
it's only one of the ways, but I think it may help here) is to look at
each node like this: Current _spills_ away from a node based on the
voltage at the node and each of the conductances away from it and
current _spills_ back into a node based on the voltages at the nearby
nodes and the conductances back in.
So when I wrote:
V(2)*(G1+G2+G3)V(3)*G3V(1)*G1V(4)*G2 = 0
I was thinking,
"Hmm. V(2) will spill current away based on the sum of the
conductances leaving the node. This is G1, G2, and G3, so the current
spilling away is simply V(2)*(G1+G2+G3). But, current is spilling
into the node based on the three different voltages at the nearby
nodes coming back through these same conductances, so this amount will
be of opposite sign and will be V(1)*G1, (V3)*G3, and V(4)*G2.
Since the total current arriving into a node must be the same as the
total current leaving it, the sum of these two must be zero  or else
we are in for big trouble!"
Similar thinking gets you the results for V(3), too. Later on, I'll
apply this thinking to the node at V(1) [V(4) would work, as well] in
order to compute the total current.
Okay, back to reality. Since V(4)=0 and V(1)=V, the above reduces
slightly to:
V(2)*(G1+G2+G3)V(3)*G3V*G1 = 0
V(3)*(G3+G4+G5)V(2)*G3V*G4 = 0
In the above pair, you have two unknowns. These are V(2) and V(3).
You also have two equations, luckily. So let's rewrite them into
slightly different form which calls out the constants a little more
clearly:
V(2)*[G1+G2+G3] + V(3)*[G3] = [V*G1]
V(2)*[G3] + V(3)*[G3+G4+G5] = [V*G4]
which can be rewritten as:
a*x + b*y = c
d*x + e*y = f
where x is V(2) and y is V(3) and with the obvious substitutions for
a, b, c, d, e, and f.
The solution to such a pair of equations can be done through simple
algebraic manipulation, but it is often shown through setting up a
basic matrix form because it is easy to visualize without getting too
caught up in the detailed manipulations:
[ a b ] [x] [c]
[ ] * [ ] = [ ]
[ d e ] [y] [f]
The solution is, of course:
[ a b ]1 [c] [x]
[ ] * [ ] = [ ]
[ d e ] [f] [y]
So you just need to compute:
c*e  f*d (G1*(G3+G4+G5) + G3*G4)
V(2) = x =  = V * 
a*e  b*d (G1+G2+G3)*(G3+G4+G5)  G3^2
and,
f*a  c*d (G4*(G1+G2+G3) + G3*G1)
V(3) = y =  = V * 
a*e  b*d (G1+G2+G3)*(G3+G4+G5)  G3^2
To solve for the voltages at the two nodes. After you have the
voltages, you should be able to find all the currents, of course.
You can look up matrix solutions in most any precalculus (preparation
for linear systems) or decent algebra math book (solving intersections
of two different lines is a very common task in algebra 2, I think.)
Or you can work through the solution of a twoline intersection
problem, keeping the terms abstract, and see it just as clearly.
(You might also want to visualize the actual "lines" involved and how
they relate back to the diagram, but that's for another time.)
Anyway, what about the total current through the system? Well, you
can apply the same reasoning I was applying before, to think like
this:
"The total current flowing into node V(1) from the voltage source
would give me what I need to know. So let's do up an equation for
that node. The total current spilling into the node from the voltage
source must be equal to the net of what happens when I only consider
the other connections. So, current spilling out through R1 and R4 are
simply V(1)*(G1+G4) and spilling back in through those resistances
must be V(2)*G1 and V(3)*G4. What arrives through the only other
connection, which is from the voltage source, must be this amount.
So: I(total) = V(1)*(G1+G4)  V(2)*G1  V(3)*G4"
Since you already now have V(2) and V(3), computing this becomes easy.
Let's select some exact values for the resistors.
R1 = 1200
R2 = 2700
R3 = 330
R4 = 1800
R5 = 680
and,
V = V(1) = 12V.
If you do your equations as I did (and if I didn't get all of this
dead wrong), it should result in:
V(2) = 5.7881 V
V(3) = 4.7872 V
And the total voltage source current is 9.184 mA. Net resistance of
the bridge is about 1306.66 ohms.
Jon