# Completely Lost in Translation/Simplification

• Jan 4th 2010, 02:38 PM
dkaksl
Completely Lost in Translation/Simplification
Ok, check this out. I have the following information;

$R'=\frac{V}{I}$

$I=\frac{V}{R}+\frac{V}{r_v}$

Taking $I$ and plugging it into $R'$, the following happens;

$R'=\frac{V}{\frac{V}{R}+\frac{V}{r_v}}=\frac{V}{\f rac{V}{R+r_v}}=\frac{V(R+r_v)}{V}=R+r_v$

So far so good? Well, $R+r_v$ is the same as $\frac{Rr_v}{R+r_v}$ right? I mean, $\frac{Rr_v}{R}+\frac{Rr_v}{r_v}$ once simplified, gives $R+r_v$ right?

The problem I have asks to show what $\frac{|R'-R|}{R}$ is in $R$ and $r_v$.

Solving looks like this; $\frac{|R'-R|}{R}=\frac{|\frac{Rr_v}{R+r_v}-R|}{R}=\frac{|\frac{Rr_v-R(R+r_v)}{R+r_v}|}{R}=\frac{|Rr_v-R(R+r_v)|}{R(R+r_v)}=\frac{|r_v-(R+r_v)|}{R+r_v}$

$=\frac{|-R|}{R+r_v}=\color{red}\frac{R}{R+r_v}$

However, if I plug in $R'=R+r_v$ instead of $R'=\frac{Rr_v}{R+r_v}$ the following ensues;

$\frac{|R'-R|}{R}=\frac{|(R+r_v)-R|}{R}=\color{red}\frac{r_v}{R}\neq{\frac{R}{R+r_v }}$ right?

Or can someone show me that $\frac{r_v}{R}=\frac{R}{R+r_v}$ because that would be AWESOME.
• Jan 4th 2010, 03:06 PM
Can you double-check....

$\frac{V}{R}+\frac{V}{r_v}=\frac{r_v}{r_v}\ \frac{V}{R}\ +\frac{R}{R}\ \frac{V}{r_v}$
• Jan 4th 2010, 03:09 PM
dkaksl
Double check what?
• Jan 4th 2010, 03:12 PM
skeeter
Quote:

Originally Posted by dkaksl
Ok, check this out. I have the following information;

$R'=\frac{V}{I}$

$I=\frac{V}{R}+\frac{V}{r_v}$

Taking $I$ and plugging it into $R'$, the following happens;

$R'=\frac{V}{\textcolor{red}{\frac{V}{R}+\frac{V}{r _v}}}=\frac{V}{\textcolor{red}{\frac{V}{R+r_v}}}= ...$

So far so good?

no ...

$\textcolor{red}{\frac{V}{R} + \frac{V}{r_v} \ne \frac{V}{R + r_v}}$

$\frac{V}{R} + \frac{V}{r_v} = \frac{Vr_v}{Rr_v} + \frac{VR}{Rr_v} = \frac{V(r_v+R)}{Rr_v}$

so ...

$\frac{V}{\frac{V(r_v+R)}{Rr_v}} = \frac{Rr_v}{r_v+R}$
• Jan 4th 2010, 03:15 PM
dkaksl
Thanks guys. I did not know my maths as well as I thought I did.