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

2. Originally Posted by oneway1225
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
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.

3. Originally Posted by oneway1225
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
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?