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Why does X/XA square root become square root of XR Ohms calculation

While explaining Ohms law to someone I realised I did not understand why http://mathhelpforum.com/image/png;b...AASUVORK5CYII= becomes http://mathhelpforum.com/image/png;b...AASUVORK5CYII=

I know , using practical numbers but I do not know how to prove it

Re: Why does X/XA square root become square root of XR Ohms calculation

Quote:

Originally Posted by

**TeslaCoil**

$\displaystyle \sqrt {\frac{P}{R}} = \frac{{\sqrt {PR} }}{R}$

Thus $\displaystyle \frac{P}{{\sqrt {PR} /R}} = \frac{{PR}}{{\sqrt {PR} }} = \sqrt {PR} $

Re: Why does X/XA square root become square root of XR Ohms calculation

Your "E" is usually written as "U".

You know $\displaystyle E=\frac{P}{\sqrt{\frac{P}{R}}}$? In this case, the derivation is simple and can be done like this:

$\displaystyle E=\frac{P}{\sqrt{\frac{P}{R}}} = \frac{P}{\frac{\sqrt{P}}{\sqrt{R}}} = \frac{P \sqrt{R}}{\sqrt{P}} = \frac{\sqrt{P}\sqrt{P} \sqrt{R}}{\sqrt{P}} = \sqrt{P}\sqrt{R}=\sqrt{PR}$

Edit: Too slow.

Re: Why does X/XA square root become square root of XR Ohms calculation

Hello, TeslaCoil!

Quote:

$\displaystyle \text{Why does }\,\frac{P}{\sqrt{\frac{P}{R}}} \;=\;\sqrt{PR}\,?$

Can you handle fractional exponents?

$\displaystyle \frac{P}{\sqrt{\frac{P}{R}}} \;=\;\frac{P}{\left(\frac{P}{R}\right)^{\frac{1}{2 }}} \;=\;\frac{P}{\frac{P^{\frac{1}{2}}}{R^\frac{1}{2} }}} \;=\;P\cdot\frac{R^{\frac{1}{2}}}{P^{\frac{1}{2}}} \;=\;P^{\frac{1}{2}}\cdot R^{\frac{1}{2}} \;=\; (PR)^{\frac{1}{2}} \;=\;\sqrt{PR} $

Re: Why does X/XA square root become square root of XR Ohms calculation

Thanks very much and yes! I do remember doing exponents ( 1980's)

I am OK with the explanation __except where the fourth expression inverts P&R __- why does that work . I am OK with the fifth , sixth and seventh expressions

How does every one put mathematical symbols in here? Math pad and cut and past ?

BTW I was in Lexington MA. last October -nice area

Re: Why does X/XA square root become square root of XR Ohms calculation

Thank you

I was OK with the explanation until the R &P expressions were inverted - how does that work ... using practical figure I see it does but why

Sorry for using E instead of the SI unit V but the old "ohms magic circle" I was using for my friend showed E as ( energy) for volts and I did not want to confuse him at this point

The blind leading the blind comes to mind