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I've been learning about Ohm's and Watt's law throughout this chapter so I'm already familiar with substituting Ohm's Law for part of Watt's law to get values. But nowhere in the chapter did it go over getting the ohmic value of the resistor. As far as I can tell $$ P = \frac{V^{2}}{R} $$ is the formula that I'll need to get it. I think my issue is basic algebra knowledge, how to get the V^2 to the other side. After trying to solve this the first time I checked my answer and realized I did it wrong and already know the answer is 4Ω :/ So, that kind of tells me it's someting like 2^2 V times 1 W. But I want to know how to do it correctly using the math. If I need ot show what I've tried already I can but using Latex is tough for me and takes a lot of time to write out the problem. I will if I need to though.

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You have the right equation - now just solve for $R$. This means you multiply both sides by $R$, and divide both sides by $P$. In steps, you started with

$$P = \frac{V^2}{R}$$

Multiply both sides by $R$:

$$P\ R = \frac{V^2}{R}R = V^2$$

Now divide both sides by $P$, to get

$$\frac{P}{P}R = \frac{V^2}{P}\\ R = \frac{V^2}{P}$$

Physics equations manipulate just like other equations...

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  • $\begingroup$ Nicely done, just got it in before I finished my answer - lol $\endgroup$
    – user60063
    Commented Oct 10, 2014 at 22:03
  • $\begingroup$ OMG, I feel so dumb. LOL I was making it WAY to hard, mainly because I've been square rooting both sides in the last couple of problems. Thanks for the reply, I appreciate it...maybe I've been studying too long. $\endgroup$
    – echo
    Commented Oct 10, 2014 at 22:06
  • $\begingroup$ @CosmoCosmo - sometimes you just have a simple block. Don't worry about it. It seemed from how you asked the question that you really did know the answer but it wasn't coming. $\endgroup$
    – Floris
    Commented Oct 10, 2014 at 22:10
  • $\begingroup$ @Floris Thanks for the vote of confidence, I need it after that one. hahaha. Thank you! $\endgroup$
    – echo
    Commented Oct 10, 2014 at 22:15
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    $\begingroup$ LOL, there we go! $\endgroup$
    – echo
    Commented Oct 10, 2014 at 22:20

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