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

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Homework Help Overview

The problem involves finding the minimum value of n in the context of equivalent resistances of two resistors in series and parallel configurations. The relationship between the series and parallel resistances is given, with specific equations provided for each configuration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equations for series and parallel resistances, exploring the implications of substituting p as nq. There is an attempt to derive a formula for n and a consideration of its minimum value.

Discussion Status

Some participants express confusion regarding the minimum value of n, questioning the possibility of n being zero given that resistances are positive. Others suggest using derivatives and the relationship between arithmetic and geometric means to explore the problem further. A potential resolution is hinted at, but no consensus is reached.

Contextual Notes

Participants note that resistances are positive quantities, which influences the discussion about the possible values of n. There is also mention of using fixed values for one resistor to minimize n with respect to the other.

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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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