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Math Help - Completely Lost in Translation/Simplification

  1. #1
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    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.
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  2. #2
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    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}
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  3. #3
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    Double check what?
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  4. #4
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    Quote Originally Posted by dkaksl View Post
    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}
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  5. #5
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    Thanks guys. I did not know my maths as well as I thought I did.
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