What Is the Minimum Value of n for Resistors in Series and Parallel?

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The discussion revolves around finding the minimum value of n in the context of resistors in series and parallel, where the equivalent resistance in series is p and in parallel is q, with the relationship p = nq. Initial calculations suggested that n could be zero, but this was deemed impossible since resistances are positive quantities. Further analysis using the arithmetic mean-geometric mean inequality led to the conclusion that the minimum value of n is 4. The final consensus confirms that the minimum value of n, derived from the relationship between the resistances, is indeed 4.
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Homework Statement


The equivalent resistance of the series combination of two resistors is p.When
they are joined in parallel, the equivalent resistance is q. If p = nq, find the
minimum possible value of n.

Homework Equations


1/Rparallel = 1/R1+1/R2+...
Rseries = R1+R2+...

The Attempt at a Solution



R1+R2=p
R1 R2 / ( R1 + R2 ) = q

Then, I substitute p as nq

R1+R2=nq
R1 R2 / ( R1 + R2 ) = q

I divide the equation 1 and equation 2 and get
##n=\frac{(R_1+R_2)^{2}}{R_1R_2}##
which I think that the minimum value is zero

However, the textbook says that the answer is 4
Please help
 
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terryds said:

Homework Statement


The equivalent resistance of the series combination of two resistors is p.When
they are joined in parallel, the equivalent resistance is q. If p = nq, find the
minimum possible value of n.

Homework Equations


1/Rparallel = 1/R1+1/R2+...
Rseries = R1+R2+...

The Attempt at a Solution



R1+R2=p
R1 R2 / ( R1 + R2 ) = q

Then, I substitute p as nq

R1+R2=nq
R1 R2 / ( R1 + R2 ) = q

I divide the equation 1 and equation 2 and get
##n=\frac{(R_1+R_2)^{2}}{R_1R_2}##
which I think that the minimum value is zero

However, the textbook says that the answer is 4
Please help

R1+R2=nq. Resistances are positive quantities. If n=0 R1+R2 =0. Is it possible?
 
ehild said:
R1+R2=nq. Resistances are positive quantities. If n=0 R1+R2 =0. Is it possible?
Hmm.. Yes, you're right.. It's impossible

So, how come we get 4 ??

I think of using derivative to get the minimum value..
The y-axis is n
But, I don't know what my x-axis is.
 
hint: fix R2 = constant, then minimize n with respect to R1 & see what if anything comes out of it.
 
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ehild said:
How are the arithmetic mean and geometric mean related? Apply that.

https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means
You can also differentiate with respect to any resistance and equate the derivative with zero.
##
n=\frac{R_1^2+2R_1R_2+R_2^2}{R_1R_2} = 2 + \frac{R_1^2}{R_1R_2} + \frac{R_2^2}{R_1R_2}
##

The AM GM of function y= ##\frac{R_1^2}{R_1R_2} + \frac{R_2^2}{R_1R_2}## is
##
\frac{R_1^2+R_2^2}{R_1R_2} \geq 2\sqrt{\frac{R_1^2 R_2^2}{R_1^2 R_2^2}}
##
## \frac{R_1^2+R_2^2}{R_1R_2} \geq 2 ## The minimum value of this function is 2
So, the minimum value of n is 2+2 = 4...
Thanks a lot for your help !
 
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