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I suspect it must be something very simple but for the life of me I cannot figure out how from the below (which models a circuit of two resistors connected in parallel) $$i=v\left(\frac{1}{R1}+\frac{1}{R2}\right)$$ we get to the following $$v=i\frac{1}{(\frac{1}{R1}+\frac{1}{R2})}$$ I suspect we get an intermittent step of $$\frac{1}{v}=\frac{1}{i}\left(\frac{1}{R1}+\frac{1}{R2}\right)$$ but how do we flip it over to obtain just the final form?

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You just reciprocate. What is $1/(1/i)$, just $i$ and the resistor values just remain in that form. – Amzoti Sep 24 '13 at 15:22
You have $a = bc$ in the first equality. You have $b = a/c$ in the second. So to get from the first to the second you divide both sides by $1/R_1 + 1/R_2$. Right? – Marek Sep 24 '13 at 15:26
If $ax=b$, then you have $x = \frac{b}{a}$. – copper.hat Sep 24 '13 at 15:29
I see now, Marek, you are spot on. I didn't realize I could do the reciprocate bit as outlined by Amzoti (as I didn't realize $\frac{1}{x}=\frac{1}{y}$ is the same as $x=y$ and also somehow I was not seeing that we can do it the other way around as you suggest Marek, that is instead of dividing by $i$ and dividing by $v$ we can go straight to dividing by $(\frac{1}{R1}+\frac{1}{R2})$, thank you so much!!! :) It has been quite a while since I have done math at school hence sometimes simple things trip me over... – dreamwalker Sep 24 '13 at 15:34
up vote 1 down vote accepted

As it turns out I can obtain the final form by dividing both sides of $$i=v(\frac{1}{R1}+\frac{1}{R2})$$ by $$(\frac{1}{R1}+\frac{1}{R2})$$ or alternatively if I start with dividing both sides by $v$ and then by $i$ and obtain $$\frac{1}{v}=\frac{1}{i}\left(\frac{1}{R1}+\frac{1}{R2}\right)$$ I can just flip the numerators and denominators since if I have $$\frac{1}{a}=\frac{1}{b}$$ if I reciprocate (or flip the numerator with the denominator on both sides) the equality still holds $$a=b$$ Either way I arrive at $$v=i\frac{1}{(\frac{1}{R1}+\frac{1}{R2})}$$ Seem a bit silly now for asking the question but thank you very much for all your help! :) Somehow couldn't see this on my own.

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