Resistors in series/parallel

This is a classic problem for circuit analysis classis as the diamond shape configuration throws alot of people off. In order to determine an equivalent value, it is necessary to pick the problem apart in smaller step.

First, redraw the circuit in a more familiar form with the resistors drawn horizontally and vertically.

Once that is done, you should be able to see that the circuit is a simple series - parallel combination. Proceed through the circuit, simplifying the parallel combinations to get the single value equivalence.

2. Tim

-- Deep Fryer: a very philosophical monk. Website:

(R1 || R2) + (R3 || R4).

Or not.

Thanks for your help, I would be happy if you could now help me confirm this:

R1 = 1/2 R2 = 3/2 R3 = 1/2 R4 = 3/2 R5 = 1/2

Does Rtot = 3/4 ?

Use the PI -> WYE conversion formula for R1,R2,and R5. That simplifies it...

ra=r1*r2/(r1+r2+r3) rb=r1*r5/(r1+r2+r3) rc=r2*r5/(r1+r2+r3)

Then, r = ra + parallel(rb+r3,rc+r4)

so

r = (r1 r2 r3 + r1 r2 r4 + r1 r2 r5 + r1 r3 r4 + r2 r3 r4 + r1 r4 r5 +

r2 r3 r5 + r3 r4 r5) / (r1 r3 + r1 r4 + r2 r3 + r1 r5 + r2 r4 + r2 r5 +

r3 r5 + r4 r5)

Please,

Do your own homework... We already graduated but on our own.

Fester

"Joachim" schreef in bericht news: snipped-for-privacy@g14g2000cwa.googlegroups.com...

Joachim,

That's hard to say without knowing the values of the resistors involved. Besides, aren't you supposed to find the solution yourself?

petrus bitbyter

Yes.

This is a classic Wheastone bridge.

Look it up (Wikipedia has a good article :

Cheers

PeteS

Cute. Now, try R1/R3 != R2/R4.

The OPs homework was probably presented with these values, and he tried to derive an equation for the resistance, without realizing that R5 doesn't pass any current... The problem is a bit harder if you don't notice that fact.