Infinate Resistance Problem

Electricity and Magnetism Level 2

A series circuit is such that there are infinitely many resistors in series shown in the diagram.

Find the total resistance in $\Omega$ .

The answer is 2.

1 solution

Jacob Morris
May 10, 2018

Question credit - https://isaacphysics.org/

The trick is to spot that the resistance changes by a factor of $\frac{1}{2}$ , making this a geometric series.

N.B - For a gemotric series with first term $a$ and common difference $r$ , the sum of the first $n$ terms is given by:

${S_n } = \frac{a(r^n-1)}{1-r}$

If $|r|<1$

${S_\infty} = \frac{a}{1-r}$

Since our geometric series is in terms of $R$ we can deduce the total resistance by finding the sum approaching infinity using the above formula.

$a=1$
$r=\frac{1}{2}$

${S_\infty} = \frac{1}{1- \frac{1}{2}} = 2\Omega$

You need to change the labelling of the $n$ th resistor in the picture, it should read $2^{\text{-}n} \Omega$ or $\frac{1}{2^n} \Omega$ .

zico quintina - 3 years, 1 month ago

Hi, can you elaborate please?

Jacob Morris - 3 years, 1 month ago

If you mean the resistance of the nth resistor then it's technically (1/2)^n-1 ohms but that gives the game away so for the problems sake it's easier to leave the resistance as n.

Jacob Morris - 3 years, 1 month ago

@Jacob Morris – I understand you not wanting to give the game away, but then you should leave the label blank or call it $x \ \Omega$ . If you leave it as is, $n \ \Omega$ means that the resistance of a resistor matches its position (because underneath it says position $n$ ), i.e. the $6$ th resistor is $6 \ \Omega$ , the $13$ th resistor is $13 \ \Omega$ , etc., which is false.

zico quintina - 3 years, 1 month ago

@Zico Quintina – I agree but your formula isn't correct. I'll change it to k when I get time

Jacob Morris - 3 years, 1 month ago

Infinate Resistance Problem

The answer is 2.

1 solution

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