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Similar-looking formula

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The equivalent resistance R of a parallel circuit

can be determined by

\displaystyle \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}.

A similar-looking formula found in a basic mathematics question involving parallel lines as shown below:

\displaystyle \frac{1}{a}=\frac{1}{a_1}+\frac{1}{a_2}.

The length of the middle line segment, a, depends on the lengths of outer lines segments and is independent of the distance between them.

In a circuit, we can consider more than two resistors:

And the equivalent resistance can be determined by

\displaystyle \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

and it helps me to make an analogy between the ‘middle length’ and equivalent resistance like

\displaystyle \frac{1}{a}=\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}.

Another formula in Physics pops up in my mind:

\displaystyle \frac{1}{f}=\frac{1}{u}+\frac{1}{v}

it is the lens formula, a similar-looking formula.

For convex lens, the object and image distances are 1 and 2 respectively (say), but if the image is virtual, we need to plug v=-2 in the formula to get the focal length,

\displaystyle \frac{1}{f}=\frac{1}{1}+\frac{1}{-2}

What is the so-called insight of the adding-minus-sign move in the basic mathematics question?

Just think of the following case: the given line segments (of lengths a_1,a_2) on the opposite sides of the ‘base line’. That is

We can still apply

\displaystyle \frac{1}{a}=\frac{1}{a_1}+\frac{1}{a_2}.

e.g. a_1=2,a_2=1,a=? Taking negative length for the line segment under the base line, thus

\displaystyle \frac{1}{a}=\frac{1}{2}+\frac{1}{-1}

Yield a=-2, meaning the length is 2 and its position is under the base line.

(Refer to the figure above, we can just regard a_2 as the middle length, and we’ll have \frac{1}{a_2}=\frac{1}{a_1}+\frac{1}{a} to get rid of negative thing.)

The adding-minus-sign move can also be found when applying section formula, but I seldom use it now.

Nothing special.

Exercise
Prove all the formulae seen in this post.


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