Complex resistor divider

Use Thevenins Theorem.

Lets call them R1 (L), R2 (R) and R3 (lower).

For no particular reason, start with R1 & R3:

Vth = 15V*R3/(R1+R3)

Rth = R1*R3/(R1+R3) = R1//R3 where // = parallel combination

Then you have:

Vth ---[Rth]--x--[R2]-- +5V

Now use the principle of superposition:

Vx = Vx(Vth = 0) + Vx(+5V = 0)

Vx(Vth = 0) = +5V*Rth/(R2 + Rth)

Vx(+5V = 0) = Vth*R2/(R2+Rth)

with R1 = R2 = R3 = 1k, we have:

Vth = 7.5V

Rth = 0.5k

Vx(Vth = 0) = 5/3V

Vx(+5V = 0) = 7.5*2/3 = 5V

so Vx = 6.667V (too lazy to write infinite number of 6's)

There are plenty of other ways of solving this too.

HTH

Cheers Terry

Probably a sci.electronics.basics question, but....

The way I like to look at this, more generally, as:

R1 Vx? R2 V1 ---/\\/\\---+---/\\/\\--- V2 | | / \\ R3 / | | V3

V1*R2*R3 + V2*R1*R3 + V3*R1*R2 The general answer is: ------------------------------ = Vx R1*R2 + R2*R3 + R1*R3

In other words, the sum of every voltage times the resistors it isn't connected to, summed up and then divided by the sum of every product combination of two resistors.

I can think of two different ways off the top of my head to see this result.

(1) Sum of currents to the Vx node must be zero.

In this, you set up three equations about the current flows. You don't know what Vx is, just yet, but whatever it is these represent the currents flowing into the node. (They are consistently __into__ [or __outof__] because I consistently use Vx-Vn.)

I(R1) = (Vx - V1) / R1 I(R2) = (Vx - V2) / R2 I(R3) = (Vx - V3) / R3

These must all sum to zero. In other words, the node cannot just sit there accumulating electrons (or else a huge disaster would happen.) All currents must sum to zero this way:

I(R1) + I(R2) + I(R3) = 0

Now, you can replace them:

(Vx - V1) / R1 + (Vx - V2) / R2 + (Vx - V3) / R3 = 0

Multiply both sides by R1*R2*R3, to eliminate all the divisors:

R2*R3*(Vx - V1) + R1*R3*(Vx - V2) + R1*R2*(Vx - V3) = 0

You can extract and adjust this to:

R2*R3*Vx + R1*R3*Vx + R1*R2*Vx = R2*R3*V1 + R1*R3*V2 + R1*R2*V3

or,

Vx * (R2*R3 + R1*R3 + R1*R2) = R2*R3*V1 + R1*R3*V2 + R1*R2*V3

which becomes what I wrote above.

(2) See things as "spilling" inward and outward through conductances, at the same time.

This is more a 'spice' way of looking at it. The equation for node two is set up as:

-G1*V1 + -G2*V2 + -G3*V3 + Vx*(G1+G2+G3) = 0

The G-values are just the conductances for each R, instead of their resistances. In other words, G1=1/R1, G2=1/R2, and G3=1/R3. So one way of reading this equation is this way -- V1 spills through conductance G1 into node Vx, V2 spills through conductance G2 into node Vx, V3 spills through conductance G3 into node Vx; while at the same time, Vx spills back outward through conductances G1, G2, and G3, all at the same time. And the sum of these 'spills' must be a net zero. In other words, Vx must spill away just as much current as spills in from the surrounding nodes.

This equation is then, just:

Vx * (G1+G2+G3) = G1*V1 + G2*V2 + G3*V3

So,

Vx = (G1*V1 + G2*V2 + G3*V3) / (G1+G2+G3)

Which, substituting in resistors is just:

V1/R1 + V2/R2 + V3/R3 Vx = --------------------- 1/R1 + 1/R2 + 1/R3

That may actually be a more helpful equation for you to think about. But it can be converted to the first one by multiplying by 1, but a value of 1 formed by (R1*R2*R3)/(R1*R2*R3):

V1/R1 + V2/R2 + V3/R3 R1*R2*R3 Vx = --------------------- * -------- 1/R1 + 1/R2 + 1/R3 R1*R2*R3

That will get you the first equation. So you can see how to get back and forth, this way.

Plugging in R1=R2=R3=1k and V1=15, V2=5V, V3=0, you get:

Vx = (15*1k*1k+5*1k+1k)/(1k*1k+1k*1k+1k*1k) = (20*1k*1k)/(3*1k*1k)

or,

Hope that helps.

Jon

I meant, "node Vx" not "node two." Sorry about that.

Jon

On a sunny day (Sat, 20 Jan 2007 18:55:54 +0800) it happened budgie wrote in :

very nice !!

The simplest way is to apply symmetry (just like the cube of resistors..... )

(15+5+0)/3 = 6.6666667V

Indeed. Kinda cheating though, as the OP made life trivial by picking all R's equal.

Well thats 3 different approaches so far. any more takers?

Cheers Terry

(cut)

Very helpful, thanks John.

You could use superposition. Ground the 5v supply, get 5.00 at X. Fix it and ground the 15v supply, get 1.666. Add.

That scheme will work for any resistor values.

John

Took me a minute to put in to LTSpice and compute the operating point.

Sorry I didn't want to wade through the math.

Joe

Call now...Joe's Spice Calculation Service only $100.00 per node!!! Thurday special...buy 25 nodes..get one free! :) D from BC

(15V/R1 + 0V/R2 + 5V/R3) * (R1||R2||R3) = 20V/R * R/3 = 6.6667 V

Works for any voltages and resistances

Best regards, Spehro Pefhany