# Reciprocals [Solved!]

**stephenB** 22 Dec 2015, 05:37

### My question

I am doing some problems with amateur radio where they ask for the "reciprocal of reciprocals" . What is actually happening when you look for a reciprocal? Thank you.

### Relevant page

3. Division of Algebraic Expressions

### What I've done so far

I understand how to find the reciprocal, but I don't understand what it is that I am doing.

X

I am doing some problems with amateur radio where they ask for the "reciprocal of reciprocals" . What is actually happening when you look for a reciprocal? Thank you.

Relevant page
<a href="/basic-algebra/3-division-algebra.php">3. Division of Algebraic Expressions</a>
What I've done so far
I understand how to find the reciprocal, but I don't understand what it is that I am doing.

## Re: Reciprocals

**Murray** 23 Dec 2015, 09:12

Hi stephenB

I suspect you are talking about total resistance in a parallel circuit, right?

The total resistance `RT` for 3 resistors `R1`, `R2`, and `R3` in parallel is given by

`1/RT = 1/R1 + 1/R2 + 1/R3`

This actually stumps a lot of people, since it seems counter-intuitive, and I think that sets up a block.

We need to add the right side as follows:

`(R2R3 + R1R3 + R1R2 ) / (R1R2R3)`

Now, since this is `1/RT` (the left hand side), and we want `RT`, we need to find the reciprocal of both sides.

`RT = (R1R2R3) / (R2R3 + R1R3 + R1R2)`

A "reciprocal of a reciprocal" just gets you back to what you started with.

**Example:** Reciprocal of `2` is `1/2`, then the reciprocal of `1/2` is `2`. But that's not exactly what we are doing in the resistors problem.

More precisely, we are adding the reciprocals of the individual resistances, then taking the reciprocal of the result.

Hope that makes some sense.

X

Hi stephenB
I suspect you are talking about total resistance in a parallel circuit, right?
The total resistance `RT` for 3 resistors `R1`, `R2`, and `R3` in parallel is given by
`1/RT = 1/R1 + 1/R2 + 1/R3`
This actually stumps a lot of people, since it seems counter-intuitive, and I think that sets up a block.
We need to add the right side as follows:
`(R2R3 + R1R3 + R1R2 ) / (R1R2R3)`
Now, since this is `1/RT` (the left hand side), and we want `RT`, we need to find the reciprocal of both sides.
`RT = (R1R2R3) / (R2R3 + R1R3 + R1R2)`
A "reciprocal of a reciprocal" just gets you back to what you started with.
<b>Example:</b> Reciprocal of `2` is `1/2`, then the reciprocal of `1/2` is `2`. But that's not exactly what we are doing in the resistors problem.
More precisely, we are adding the reciprocals of the individual resistances, then taking the reciprocal of the result.
Hope that makes some sense.

## Re: Reciprocals

**stephenB** 24 Dec 2015, 06:28

Got it - thanks.

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