# Resistors in series/parallel

Discussion in 'Electronic Basics' started by Joachim, Nov 14, 2005.

3. ### John - kd5yiGuest

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

Or not.

4. ### Noway2Guest

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.

5. ### JoachimGuest

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 ?

Fester

7. ### Bob MonsenGuest

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)

---
Regards,
Bob Monsen

One cannot inquire into the foundations and nature of mathematics without
delving into the question of the operations by which the mathematical
activity of the mind is conducted. If one failed to take that into account,
then one would be left studying only the language in which mathematics is
represented rather than the essence of mathematics.
- Luitzen Brouwer

8. ### petrus bitbyterGuest

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.

11. ### Bob MonsenGuest

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.  