i don't know how to explain this

A relationship between power P, current I, and resistance R in an electric circuit is given by the equation

R=P/I^2

Explain the effect a change in current has on the resistance

Printable View

- Apr 8th 2009, 05:59 AMoneway1225Please explain this
i don't know how to explain this

A relationship between power P, current I, and resistance R in an electric circuit is given by the equation

R=P/I^2

Explain the effect a change in current has on the resistance - Apr 8th 2009, 06:33 AMmasters
Hi oneway1225,

$\displaystyle R=\frac{P}{I^2}$

According to your formula,**R**varies inversely as the square of**I**.

This means as**I**increases,**R**decreases and as**R**decreases,**I**increases.

The current flowing in an electrical circuit at a constant potential varies inversely as the resistance of the circuit. - Apr 8th 2009, 07:02 AMstapel
To learn the "lingo" of "variation" equations, try

**here**.

To "see" the effect in this particular exercise, try plugging numbers in. For instance, if the current I is doubled, so you're plugging "2I" in where you'd had just "I", what happens?

. . . . .$\displaystyle R_{new}\, =\, \frac{P}{(2I)^2}\, =\, \frac{P}{4I^2}\, =\, \frac{1}{4}\left(\frac{P}{I^2}\right)$

Since the original R had been P/I^2, then:

. . . . .$\displaystyle R_{new}\, =\, \frac{1}{4}\left(R_{orig.}\right)$

In doubling the current, the resistance is reduced to one-fourth.

Now what happens if the current is halved? (Wink)