# Resistor equation problem

1. Sep 26, 2015

### terryds

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
1/Rparallel = 1/R1+1/R2+...
Rseries = R1+R2+...

3. 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

2. Sep 27, 2015

### ehild

R1+R2=nq. Resistances are positive quantities. If n=0 R1+R2 =0. Is it possible???

3. Sep 27, 2015

### terryds

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.

4. Sep 27, 2015

### ehild

5. Sep 27, 2015

### rude man

hint: fix R2 = constant, then minimize n with respect to R1 & see what if anything comes out of it.

6. Sep 27, 2015

### terryds

$n=\frac{R_1^2+2R_1R_2+R_2^2}{R_1R_2} = 2 + \frac{R_1^2}{R_1R_2} + \frac{R_2^2}{R_1R_2}$

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 !