Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question has to do with the concept of resistors, given two resistors in parallel the equivalent resistance is always lower than the smallest individual resistance, I am trying to convince myself that this is true.

The equivalent resistance is given by $z$ and the individual resistances are $x$ and $y$.

share|improve this question
    
Please use parentheses. $xy/x+y=(xy/x)+y$ which is not what you meant. –  Ross Millikan Sep 30 '12 at 22:37
    
suppose you write it $$\frac{1}{z} = \frac{1}{x} + \frac{1}{y}$$ and try to show $$\frac{1}{z} > \frac{1}{x}$$ and so on... –  GEdgar Sep 30 '12 at 22:41

4 Answers 4

up vote 4 down vote accepted

Note that $$z={xy\over x+y} = x\cdot {y\over x+y}$$ and that ${y \over x+y}<1$ since the numerator is smaller than the denominator.

share|improve this answer
    
Jeez such a simple explanation, can't believe I did not realize it. Thanks, will choose your answer as soon as it lets me. –  user1084113 Sep 30 '12 at 22:46

Because if $0<a<b$ then $\displaystyle\frac ca>\frac cb$. So, $$\frac{xy}{x+y}<\frac{xy}x$$

share|improve this answer

$x,y>0$ implies $$z=\frac{xy}{x+y}=\frac x{x+y}\cdot y<\frac{x+y}{x+y}\cdot y=y$$ and the same with $x$, $y$ interchanged.

share|improve this answer

I feel like I am thinking too complicated sometimes. I had proven it like that:

$$z=\frac{xy}{x+y}=\frac{xy}{x\cdot (1+\frac{y}{x})}=\frac{y}{1+\frac{y}{x}}$$

With $x,y>0$ follows

$$1+\frac{y}{x}>1\rightarrow\frac{y}{1+\frac{y}{x}}<y$$

This makes $z<y$ and $z<x$ respectively.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.