OT Working out Parallel resistor value

[zikkimalambo]

I can prolly do what I want simply by adding a resistor in series.

(3.3K +333 Ohms in series gives approx 3.6 K yes?)

However, lets assume for a moment that i've been locked in a tall tower by an evil tyrant dual combi boiler monster who will not release me until i've gotten as near as possible to 3.6K using the existing 3.3K and another resistor in parallel.

I had a look on google and it tells me (someone who once posted on usenet tells me) that in order to work out the resistance of two restors in parallel, you multiply the two values, then add the two values, then divide the output of the multiplication by the output of the addition.

However, how can you work it backwards, if you have the value of one resistor and the total resistance you are aiming at, but want to know what value to use for the other?

[manatbandq]

The normal way of stating the formula is using the reciprocals of the values. Given R1 and R2 we want to know the parallel value Rp:

1/Rp = 1/R1 + 1/R2

multiply up by Rp, R1 and R2: R1.R2 = Rp.R2 + Rp.R1 - (eqn 1)

Collect the Rp terms and divide gives, as you say: Rp = R1.R2/(R1+R2)

Now, knowing Rp, we want to find R2. Start at eqn 1 and simply collect the R2 terms together: R2(R1 - Rp) = Rp.R1

Divide, and hey presto: R2 = Rp.R1/(R1 - Rp)

What you have to remember is that resistors don't come in any old value so you're almost always going to end up with a compromise or a trimmer.

MBQ

[Andy Wade]

In ohms 3300 + 333 = 3633, or 3.633k (aka 3k633). Or two 1k8's in series gives you an exact solution.

That's impossible since a parallel resistor will only lower the value. You need a series one to raise it. You could make 3k6 from 3k9 in parallel with 47k (see formula below).

Doesn't anyone do elementary algebra at school any more?

If R1 and R2 are the two resistors and Rp is their value in parallel, then

1/Rp = 1/R1 + 1/R2, hence

R1 * R2 Rp = --------- (R1 + R2)

And

R1 * Rp R2 = --------- (R1 - Rp)

You do realise that 3k6 is a preferred value (E24 series)?

HTH

[Alan Holmes]

Perhaps he doesn't understand the difference between 'series' and 'parallel'!

[Grunff]

You can never make 3.6k by paralleling a resistor with a 3.3k one. Ever. The value will always be lower than the lowest resistor in the parallel set.

[EricP]

Use the formula

1/R1 + 1/R2 = 1/R3

R1 and R2 being your resistors and R3 being the answer

[sm_jamieson]

Yep. Another resistor in parallel is an extra path for electrons, so the resistance must be lower. You have a wider "pipe". Imagine using two hosepipes in parallel to offer *more* resistance to the water. Impossible. Simon.

[Rob Morley]

Then you're screwed, because placing more resistance in parallel will only reduce the current value.

I thought it was more like 1/R1 + 1/R2 = 1/RT

so 1/(1/RT - 1/R1) = R2

[Chip]

On 30 Jun 2005 04:55:03 -0700,it is alleged that [e-mail] spake thusly in uk.d-i-y:

I never did anything that complex :-) I use this formula (on google groups, you may need to switch to a monospaced or non-proportional font)

For resistors in parallel:

1 1 1

--------- = --- + --- R (total) R 1 R 2

Continue as necessary for other resistors. (R3 etc)

By rearranging that, if you know R1 and R(total) the formula for working out R2 would be:

1 1 1

--- = --------- - --- R 2 R (total) R 1

Any calculator will then allow you to get the values themselves by dividing 1 by the results :-)

Your example however is unworkable, if you need 3.6k, and have 3.3k, then 3.3k is as close as you're going to get if you can't use series combinations, given the absence of resistors with negative resistance values .

Putting anything in parallel with it is only going to *lower* the resistance.

HTH

[news]

Rob Morley wrote: : I thought it was more like 1/R1 + 1/R2 = 1/RT

The two formulae are exactly equivalent:

1/R1 + 1/R2 = 1/RT

multiply all terms by R1R2:

R1R2/R1 + R1R2/R2 = R1R2/RT

simplify:

R2 + R1 = R1R2/RT

rearrange:

RT = R1R2/(R1+R2)

QED!

Richard.

reply by email change 'news' to my forename.

[zikkimalambo]

All sorts of people pointed out that the reisitance of two resistors in parallel will always be lower than either one of them.

Bah!

Oh well, it's oft been said that the only stoopid question is the one you don't ask!

Thanks all.

[Dave Plowman (News)]

To avoid stretching the brain if you do a search on 'resistance calculator' or similar, I'm sure you'll come up with several freebie progs that will do this for you.

[Andrew Mawson]

Resistors in parallel if I remember my 'O' level physics are:

1/R(tot) = 1/R(1) + 1/R(2) ....... + 1/R(n)

so if two 1 ohm resuistors are used:

1/R = 1/1 + 1/1 =2

therefore 1/R = 2 therefore R = 0.5 ohms

Q.E.D.

AWEM

[Rob Morley]

I know, but I find the way I wrote it much easier to remember, just as V = IR is easier to remember than R = V/I

[Chris McBrien]

Simply put...... The product divided by the sum....

For two resistors laying side by side in parallel.

x-- 100 Proof News -

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[John]

No way in this reality are you going to "increase" the resistance by putting something in parallel with it. Now if you want to put 0,3k is SERIES then yes

[mike ring]

[e-mail] wrote in news:1120135637.469583.218910 @z14g2000cwz.googlegroups.com:

But a remarkable lot didn't spot the tripwire. :-)

I did have the unworthy thought that you were indulging in a little light trolling .... but slap my leg

[mungoh]

Many moons ago someone posed the puzzle that if you built a cube shape using one ohm resistors as the edges of the wire cube, what would be the resistance across the longest diagonal? I'm sure that a Google would turn up the page with the answer...

Mungo

[mike ring]

[e-mail] wrote in news:1120162655.630554.86880 @g47g2000cwa.googlegroups.com:

was going to explain it all myself, as I did it years ago, but I couldn't recomember how)

mike

[Ed Sirett]

This is probably hard to work out using brute force methods. Where you would have to work out the voltages on all the corners.

However by an appeal to symmetry: there are three paths from the first corner to three 'identical' corners, then six paths to the another set of 'identical' corners and finally three paths to the last corner.

So 1/3 + 1/6 + 1/3 = 5/6.

However just one resistor being different would make this very hard (=impossible for me) to work out.