Results 1 to 3 of 3

Math Help - Please explain this

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    13

    Please 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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by oneway1225 View Post
    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,

    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    Quote Originally Posted by oneway1225 View Post
    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?

    . . . . . 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:

    . . . . . 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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Explain this!
    Posted in the Math Challenge Problems Forum
    Replies: 4
    Last Post: December 30th 2010, 03:41 PM
  2. How can i explain???
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: June 6th 2010, 04:37 PM
  3. Can someone explain this to me?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 29th 2009, 08:11 AM
  4. Can someone explain how I do this one?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 15th 2008, 08:22 PM
  5. Can you explain to me how to do this?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 8th 2008, 03:55 AM

Search Tags


/mathhelpforum @mathhelpforum