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Is the equivalent resistance across a set of n resistors connected in series greater than the equivalent resistance across them when they are connected in parallel? I want to find the relation between \$R_s\$ and \$R_p\$.

My attempt:

Let the resistance of the resistors be denoted by \$R_1, R_2, R_3,..., R_n\$ and the equivalent resistance across the series and parallel combinations of the resistors by \$R_s\$ and \$R_p\$.

\$R_s=R_1+R_2+R_3+...+R_n\$

and

\$R_p=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}}\$

From first look it may seem that \$R_p\$ is less than \$R_s\$ but then on careful inspection it is not possible to ascertain which one is greater.

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6 Answers 6

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Let's try a very simple approach.

The collection of resistors will have a maximum value resistor \$R_{max}\$, and a minimum value resistor \$R_{min}\$. By definition $$R_{min} <= R_{max}$$ the equality is true if all resistors are equal, otherwise the inequality is true

The series connection will contain \$R_{max}\$, plus at least one other series resistor, which will increase its resistance, so $$R_s > R_{max}$$

The parallel connection will contain \$R_{min}\$, plus at least one other parallel resistor which will decrease its resistance, so $$R_p < R_{min}$$

as $$R_p < R_{min} <= R_{max} < R_s$$ therefore $$R_p < R_s$$

Notes

(1) We assume all the resistances are non-negative. This is true if they are all passive. There are active networks that can exhibit a negative resistance, but they are not relevant to this question.

(2) Resistance is voltage/current. The total resistance of any two resistors in series will have a resistance greater than either component. The total resistor will have a voltage equal to the sum of that over each component, with all currents being equal.

(3) The total resistance of any two resistors in parallel will have a resistance lower than either component. The total resistor will have a current equal to the sum of that through each component, with all voltages being equal.

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  • \$\begingroup\$ The other resistor in the parallel combination has greater resistance. So shouldn't it increase the resistance? \$R_p=\frac{1}{\frac{1}{R_{min}}+\frac{1}{R_{max}}}\$. There are two inversions. On the first inversion, 1/Rmin has greater value and on second inversion it reduces the equivalent resistance. \$\endgroup\$
    – MrAP
    Commented Mar 3, 2017 at 9:04
  • \$\begingroup\$ @MrAP two resistors in parallel form a total resistance that's smaller than either of them. Look at it as conductance. Two conductances in parallel form a total conductance that is larger than either of them. I must admit I thouhgt that for two reistors, it's self evident. The reciporcal of the sums of reciprocals is an excellent way to obscure what's going on. \$\endgroup\$
    – Neil_UK
    Commented Mar 3, 2017 at 9:05
  • \$\begingroup\$ I have edited my comment. \$\endgroup\$
    – MrAP
    Commented Mar 3, 2017 at 9:08
  • \$\begingroup\$ Are you saying that you can find an example where Ra and Rb in parallel have a total reesitance that is larger than min(Ra, Rb) ? \$\endgroup\$
    – Neil_UK
    Commented Mar 3, 2017 at 9:09
  • \$\begingroup\$ How do you justify that the equivalent resistance across a set of resistors connected in parallel is less than each of the resistors? \$\endgroup\$
    – MrAP
    Commented Mar 3, 2017 at 10:15
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Let's see if this works.

Start with \$n=2\$:

$$ R_s = R_1+R_2\\ R_p = \frac{R_1R_2}{R_1+R_2}\\ \text{Then: }\\ \frac{R_1R_2}{R_1+R_2} \stackrel{?}{<}R_1+R_2\\ R_1R_2 \stackrel{?}{<} (R_1+R_2)^2 $$ The last row can be derived from the one before if \$R_1+R_2>0\$ and it is true if both resistance are positive separately too, which is something ok to ask to a resistance. \$n=2\4 is our base case.

Now... induction.

Hypothesis: \$R_p[n] < R_s[n]\$

Thesis: \$R_p[n+1] < R_s[n+1]\$

We try to prove that the sequence \$R_p[n]\$ is decreasing, while \$R_s[n]\$ is increasing.

Series resistance: $$ R_s[n+1] = R_s[n] + R_{n+1} > R_s[n] \text{ if } R_{n+1}>0\\ R_s[n+1] > R_s[n] \textit{ q.e.d.} $$

Parallel resistance: $$ R_p[n+1] = \frac{R_p[n]R_{n+1}}{R_p[n]+R_{n+1}}\stackrel{?}{<}R_p[n]\\ R_p[n]R_{n+1}\stackrel{?}{<}R_p[n]^2+R_p[n]R_{n+1}\\ R_p[n]^2>0\text{ which is true, so that} R_p[n+1] < R_p[n] \textit{ q.e.d.} $$

We finally chain things up:

$$ R_p[n+1] < R_p[n] < R_s[n] < R_s[n+1] $$

It did work after all.

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  • \$\begingroup\$ Pretty good demonstration. \$\endgroup\$
    – Tagadac
    Commented Mar 2, 2017 at 10:55
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It's a purely mathematical result.

If we look at the aritmetic and harmonic means of the set of resistor values: $$ AM (R_1,\cdots, R_n) = \frac{1}{n} (R_1 + \cdots + R_n)\\ HM (R_1,\cdots, R_n) = \frac{n}{\frac{1}{R_1} + \cdots + \frac{1}{R_n}} $$

We can observe that \$R_s\$ is \$n\$ times the arithmetic mean \$AM (R_1,\cdots, R_n)\$ of the resistors, while \$R_p\$ is \$1/n\$ times the harmonic mean \$HM (R_1,\cdots, R_n)\$ of the resistors.

The Pytagoreans worked out that \$AM \geq HM\$ if the values in the set are positive (as resistors use to be). So, if \$n \geq 2\$ (at least two resistors), then:

$$ R_p = \frac{1}{n}HM (R_1,\cdots, R_n) < HM (R_1,\cdots, R_n) < AM (R_1,\cdots, R_n)< n AM (R_1,\cdots, R_n) = R_s $$

Thus, \$R_p < R_s\$ always for any \$n \geq 2\$ and positive valued \$R_1,\cdots, R_n\$.

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I think there's a simpler way to show this than the existing answers have given.

When thinking about resistors in parallel, it's often simpler to think about conductance rather than resistance. In terms of conductance, we can simply write:

$$G_p = G_1 + G_2 + \dots$$

Since all of these G's are positive numbers, we know that \$G_p\$ is greater than any of the individual \$G_n\$'s.

Therefore we know (still taking into consideration that all the values are positive) that \$R_p=1/G_p\$ is less than any of the individual \$R_n\$'s.

Since \$R_p\$ is less than any individual resistor value, it must also be less than the series combination (sum) of all the individual resistor values.

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This has been answered pretty thoroughly but I like playing around with this stuff, the min/max answer is my favourite though.

So here's an unnecessarily awkward proof just using sums. $$ R_s = \sum_{1}^{n} R_n\\ R_p = \frac{1}{\sum_{1}^{n} \frac{1}{R_n}} $$ Multiplying top and bottom of fraction in Rp by Rs: $$ R_p = \frac{R_s}{R_s\sum_{1}^{n} \frac{1}{R_n}} = \frac{R_s}{\sum_{1}^{n} \frac{\sum_{1}^{n} R_n}{R_n}} $$ Now that inner sum always contains one term that is equal to its divisor, so that can be split out and simplified to 1: $$ R_p = \frac{R_s}{\sum_{m=1}^{n} \frac{R_m}{R_m}+ \frac{\sum_{n=1}^{m-1} R_n + \sum_{n=m+1}^{n} R_n}{R_m}} = \frac{R_s}{n + \sum_{m=1}^{n} \frac{\sum_{n=1}^{m-1} R_n + \sum_{n=m+1}^{n} R_n}{R_m}}\\ R_p = \frac{R_s}{>1} $$ Now the divisor contains n, which is greater than one, and since no other part can be negative the denominator as a whole is greater than one, thus Rp will be smaller than Rs.

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If we assume that all \$R_1, R_2, \dots, R_n \$ have the equal value \$ R \$, you get the following connectivity:

\$ R_s = \sum\limits_{i=1}^n R_i = n \cdot R\$ (1)

\$ R_p = \frac{1}{\sum\limits_{i=1}^n \frac{1}{R_i}} = \frac{1}{n \cdot \frac{1}{R}} = \frac{1}{n} \cdot R \$ (2)

Resolve (1) to R => \$ R = \frac{R_s}{n} \$ plug in (2) => \$ R_p = \frac{1}{n} \cdot \frac{R_s}{n} = \frac{1}{n^2} \cdot R_s \$

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