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There are $n$ resistors connected in a parallel combination given below.

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$$\frac{1}{R_{ev}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}.......\frac{1}{R_{n}}$$

Foundation Science - Physics (class 10) by H.C. Verma states (Pg. 68)

For two resistances $R_{1}$ and $R_{2}$ connected in parallel,

$$\frac{1}{R_{ev}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{R_{1}+R_{2}}{R_{1}R_{2}}$$ $$R_{ev}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}$$

We see that the equivalent resistance in a parallel combination is less than each of the resistances.

I observe this every time I do an experiment on parallel resistors or solve a parallel combination problem. How can we prove $R_{ev}<R_{1},R_{2},R_{3},...R_{n}$ or that $R_{ev}$ is less than the Resistor $R_{min}$, which has the least resistance of all the individual resistors?

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  • $\begingroup$ I have to confess that when I saw that you had exhibited the harmonic sum form ($1/R_e = 1/R_1 + 1/R_2 + \dots$), I found the question confusing because I didn't understand what you didn't get. It's worth your time to think about this until it becomes obvious, not because this problem is so important but because it will start teaching you the habit of learning things from the functional form of the relationships that appear in your studies. $\endgroup$ – dmckee --- ex-moderator kitten Sep 27 '14 at 17:15
  • $\begingroup$ @dmckee I didn't quite get what you said. It'll probably help me if you'll elaborate. $\endgroup$ – user49111 Sep 27 '14 at 17:26
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    $\begingroup$ With experience, you will find this fact obvious, and it is worth taking the time to ponder it until it becomes so. $\endgroup$ – dmckee --- ex-moderator kitten Sep 27 '14 at 17:34
  • $\begingroup$ I rolled back this question to the original version because you shouldn't put answers, or responses to answers, in the question itself. @imakesmalltalk Feel free to post an answer of your own if you would like to offer another method of answering the question. $\endgroup$ – David Z Oct 27 '14 at 0:30
  • $\begingroup$ Once you believe the 2 resistor answer, it extends to any finite number. For each resistor, imagine combining all the rest into a single resistor by the law you cite. Now the combination of the one resistor and the combination resistor is less than the one. Do this once for each of the resistors, and you have that the total combination is less than any one. $\endgroup$ – Ross Millikan Oct 27 '14 at 2:53
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Think about current flow.
If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$ Dividing everything by the voltage give us: $$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Which is the same as: $$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Since there is more current flowing in all the resistors than through just one resistor, then the equivalent resistance must be less than the individual resistors.

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The individual resistances are all positive, so the sum $$ \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots \,,$$ is larger than the inverse of any of the individual resistances, and that means that the inverse of the sum is necessarily smaller than any of the resistances.

No mucking around with the two-resistor form required.

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We can prove it by induction. Let $$ \frac{1}{R^{(n)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n} $$ Now, when $n=2$, we find $$ \frac{1}{R^{(2)}_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_{eq}^{(2)} = \frac{R_1 R_2}{R_1+R_2} = \frac{R_1}{1+\frac{R_1}{R_2}} = \frac{R_2}{1+\frac{R_2}{R_1}} $$ Since $\frac{R_1}{R_2} > 0$, we see that $R^{(2)}_{eq} < R_1$ and $R^{(2)}_{eq} < R_2$ or equivalently $R^{(2)}_{eq} < \min(R_1, R_2)$.

Now, suppose it is true that $R^{(n)}_{eq} < \min (R_1, \cdots, R_n)$. Then, consider $$ \frac{1}{R^{(n+1)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n} + \frac{1}{R_{n+1}} = \frac{1}{R^{(n)}_{eq}} + \frac{1}{R_{n+1}} $$ Using the result from $n=2$, we find $$ R^{(n+1)}_{eq} < \min ( R_{n+1} , R^{(n)}_{eq} ) < \min ( R_{n+1} , \min (R_1, \cdots, R_n)) $$ But $$ \min ( R_{n+1} , \min (R_1, \cdots, R_n)) = \min ( R_{n+1} , R_1, \cdots, R_n) $$ Therefore $$ R^{(n+1)}_{eq} < \min ( R_1, \cdots, R_n , R_{n+1} ) $$ Thus, we have shown that the above relation holds for $n=2$, and further that whenever it holds for $n$, it also holds for $n+1$. Thus, by induction, it is true for all $n\geq2$.

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  • $\begingroup$ $\frac {R_1}{R_2} > 0$ not 1 $\endgroup$ – Omar Elawady Apr 4 '16 at 21:41
  • $\begingroup$ Your induction method is much more mathematical than the other answers....but is there any way to solve this using $AM-GM$ inequality? $\endgroup$ – Soham Jun 3 '16 at 16:11
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It might be easier to think in terms of conductance which is the inverse of resistance.

The more paths there are between A and B which conduct electricity, the greater is the amount of current which can flow - ie the greater is the conductance of the network. The total conductance is greater than that of any individual path, because each additional path always increases the amount of electricity which can be conducted, it never reduces it. In particular, the total conductance is always greater than the largest individual conductance.

Translating this back in terms of resistance R (which is the inverse of conductance S - ie R = 1/S), the total resistance is smaller than the smallest individual resistance.

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