4
$\begingroup$

Am stuck on this problem in electronics as I have ran into a bit of algebra. Am generally not too bad with algebra but I cant for the life of me solve this equation:

$V=IR$
$P=IV$

So $V = \frac{P}I$

We know that $I = \frac{V}R$

So, given that $P=IV$, rearranging $V$ to give V=$\frac{P}I$, and thus substituting I for $\frac{V}R$, $V$ can then equal: $$ V = \frac{P}{\frac{V}{R}}$$

... Right?

Well in my text books it says that v^2 =PR, and I can't for the life of me figure out how to rearrange the above equation to produce v^2=PR. I've thought multiplying V by V, maybe, to get V^2 = P/R, but of course that's wrong.

Any help would much obliged, also, is there anywhere I can go to learn algebra with multiple divisions, every maths website just does 1 division, and it seems that cross multiplying works with 1 division, but not 2?

Many thanks,

James.

$\endgroup$

2 Answers 2

4
$\begingroup$

We have:

  • $V = IR$
  • $P = IV \rightarrow I = \dfrac{P}{V}$
  • $V = IR = \dfrac{P}{V}R \rightarrow V^2 = PR$

We can derive many such relationships, see this wheel of voltage, current and power.

Update

Using your expression (I invert the fraction in the denominator and then cross multiply):

$$ V = \dfrac{P}{\dfrac{V}{R}} \rightarrow V = \dfrac{P R}{V} \rightarrow V^2 = PR$$

There are many ways to arrive at the various forms.

$\endgroup$
2
  • $\begingroup$ Maybe it does have all the needed tools, but I would imagine that a good answer takes the expression V = P/(V/R) and shows it is the same as VV=PR. $\endgroup$
    – Adam
    Nov 12, 2013 at 15:39
  • $\begingroup$ @Adam: Noted and updated. Regards $\endgroup$
    – Amzoti
    Nov 12, 2013 at 15:52
2
$\begingroup$

Since $V=IR$, we can solve for $I$ by writing $I=\frac{V}{R}$.

Also, since $P=IV$, we can solve for $I$ again by writing $I=\frac{P}{V}$.

Both equations are equal to $I$, so that means they equal each other and we have $$I =I \\ \frac{V}{R} = \frac{P}{V}$$ Then cross-multiply to get $$V^2 = PR $$ Voila!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .