# total resistance in parallel circuits

Discussion in 'Electronic Basics' started by Midnight Oil, Oct 5, 2005.

1. ### Midnight OilGuest

I am new to electronics, and I've been learning the basics of ohm's
law. I ran into the formula for finding the total resistance in a parallel
circuit, struggled with it's meaning...and I want to be sure I understand
the formula:

R(tot) = 1
-------------
1 + 1 + 1
-- -- --
R1 R2 R3

I broke the formula down like this:

R(tot) = 1 <--- E (volts)
---------------
1 + 1 + 1
-- -- -- <--- I (amps)
R1 R2 R3

In other words, the addition of 1/R1, 1/R2, and 1/R3 reveals the
current when 1 volt is applied to the circuit (E/R). Then, once we know
the current, we can divide the 1 volt by the current to reveal the total
resistance R= (E/I).

In other words, 1/Rx reveals the amount of current in one branch of the
parallel circuit, and adding these together gives us the total current in
the circuit when 1 volt is applied. If we divide 1 volt by that value, we
get the resistance in the circuit when one volt is applied.

Is my understanding of the equation correct?

I thought it was, until I read further in the book. It explained that
the reason for the 1/Rx was because it is expressing conductivity rather
than resistance.

I thought it was interesting that 2 interpretations of the formula
could co-exist...or was my own interpretation of the formula wrong?

Is it just a coincidence that the amount of conductance is equal to the
amount of current flowing when 1 volt is applied?

Thanks!

- Jamie

The Moon is Waxing Crescent (7% of Full)

2. ### Tom BiasiGuest

Yes Jamie,
It can be confusing.
If you can use your algebra ( a must in this field) you may better
understand the concept if you consider only two resistances in parallel.

With manipulation you will arrive at the - product over sum - formula. R1 x
R2/(R1 + R2) You can do two resistors (resistances), come to a conclusion,
use the result to combine the other one in the pair.

Then you will appreciate the - one over, one over formula. Yes they are
conductance's. As for you coincidence: is 2+2 the same as 2 squared? Is 4+4
the same as 4 squared?

Review the algebra.
Best Regards,
Tom

3. ### John PopelishGuest

I think so. At least you seem to be getting that an ohm is just
another way of saying 1 volt per ampere. Another way to look at the
parallel resistance formula is that you convert the resistances to
conductances (amperes per volt, as you have figured out, but now you
have a word that names that ratio). Then, after adding the
conductances together to get a total conductance you convert that
conductance back to a resistance by taking the inverse (flipping the
amperes in the numerator with the volts in the denominator to get back
to volts per ampere.
Exactly. Conductance (a bit different than conductivity, which is a
property of a bulk material) is the name for the ratio of amperes per
volt.
Not wrong, just sticking to the more fundamental units.
It is the definition of conductance.

4. ### Tim WilliamsGuest

Incorrect: the ones do not represent any units, so the only thing you'll get
out is what you put in - in this case, ohms. The intermediate step of
reciprocal resistance in ohms (which is conductance in mhos) follows from
the nature of the circuit.

Aside from the confusion on units ...
Correct.

What you are imagining is equivalent to the mathematical technique of
testing a "well-behaved" function at a convienient value like x = 1 and
extrapolating or proving other values based on this.

Given nice ohmic devices, the exact same behavior applies, as a matter of
fact, so it is true you can test and prove it in this way.
Only as I mentioned above. Gotta watch units in equations.
Nope, it's by definition in fact

Tim

5. ### redbellyGuest

That's not a bad way to interpret what's going on. But, to make the
units work out properly, we could multiply the expression by 1V/1V
(since that equals 1 and multiplying by 1 is allowed), then manipulate
things like this:

1 1V
R(tot) = ---------------- x ----
1 1 1 1V
-- + -- + --
R1 R2 R3

1V <--- E (volts)
= ----------------
1V 1V 1V
-- + -- + -- <--- I (amps)
R1 R2 R3

So each resistor has 1V across it, and has (1V/Rx) current flowing
through it.

Regards,

Mark

p.s.
Hmmmm. Years ago I had DOS version of that program. Is there
something available for Windows these days?

6. ### Pooh BearGuest

Think of it as the sum of conductances. It's simple then.

Graham

7. ### Jasen BettsGuest

you are measuring conductance if you put 1 volt across a resistor and
measure the current it passes.

I would say that both interpretations are equivalent.

Bye.
Jasen

8. ### Rich GriseGuest

I saw this derived from the sum of conductances:

G(tot) = G1 + G2 + G3

Since conductance is the reciprocal of resistance,

G(tot) = 1 / R(tot)
G1 = 1 / R1 etc

so

1 / R(tot) = (1 / R1) + (1 / R2) + (1 / R3)

Multiply through by R(tot)

1 = R(tot) * ((1 / R1) + (1 / R2) + (1 / R3))

Divide through by ((1 / R1) + (1 / R2) + (1 / R3))

1 / ((1 / R1) + (1 / R2) + (1 / R3)) = R(tot)

QED. (I've always wanted to be able to say that! )

Cheers!
Rich

9. ### Jasen BettsGuest

The Moon is Waxing Crescent (7% of Full)
it looks like the output from "pom" the bsd "phase of moon" program

The Moon is Waxing Crescent (23% of Full)

C source is available from
ftp://ftp.netbsd.org/pub/NetBSD/NetBSD-current/tar_files/src/games.tar.gz

winzip (etc) should be able to open that file, and pretty much any C compiler
should be able to produce a working executable from the source. some of the
other "games" in the package may not compile for windows as easily.

Bye.
Jasen

Today is Sweetmorn, the 62nd day of Bureaucracy in the YOLD 3171