Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Confusing relation between power,volatage and current

  1. Nov 22, 2014 #1
    Hii,i'm new in electrical and much confused bcoz of complicated relation between V,I, nd P.
    If P = VI = cons. and increasing V,decrease the value of I,
    but since P is also equal to I×I ×R and r is cons.,so decrease in I will always cause P to decrease which is supposed to remain constant.

    And i can't get this relation's(p=vi) physical or practical meaning also....
     
    Last edited: Nov 22, 2014
  2. jcsd
  3. Nov 22, 2014 #2

    Doug Huffman

    User Avatar
    Gold Member

    Practical machines are designed to a relatively narrow range of voltages (insulation, mechanical spacing) and heat dissipation capability.
     
  4. Nov 22, 2014 #3
    Thanks for reply sir,but i think,it doesn't explain, what i want to understand.
     
  5. Nov 22, 2014 #4

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    I don't know where you got the idea to inject cons (constant). Forget cons.

    P=V*I in all circumstances.

    P=I*I*R in all circumstances.

    There is no contradiction.
     
  6. Nov 22, 2014 #5
    P=V⋅I always
    P= I2⋅R when R=const.
     
  7. Nov 22, 2014 #6

    phinds

    User Avatar
    Gold Member
    2016 Award

    HUH? Where did you get that idea? Increasing V INCREASES the value of I by Ohms's Law, V=IR
     
  8. Nov 22, 2014 #7

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    Not true, P= I2⋅R even when R is variable. Just remember that we are talking about instantaneous values, not average, not RMS.
     
  9. Nov 22, 2014 #8
    And how do you define instantenous value of R?
     
  10. Nov 22, 2014 #9

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    Any way you want. Just assume any R as a function of time. Perhaps R is the resistance of a rheostat, and a three year old kid is playing with the rheostat knob. Also assume any time variation for I. P(t)=I(t)*I(t)*R(t) always.
     
  11. Nov 22, 2014 #10
    Careful there. If you define something any way you want it can lose meaning.Yes, you can define resistance as some function of time to match power dissipated but resistance is originally defined as function of v vs i. And difference in definition between static and dynamic resistance implies it's not necessarily same value for nonlinear resistor (usually it isn't).
     
  12. Nov 22, 2014 #11

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    This is crazy. V(t)= I(t)*R(t). That is the definition of Ohm's Law in real, instantaneous values.

    It works also for V(t)= I(t)*Z(t) with complex V, I, and impedance Z in AC analysis.

    Unless you are thinking of wave propagation, that is the way things are defined. What else could it be?

    We should be precise in nomenclature. I is the current through the resistance R, and V is the voltage across the two ends of R. It matters not if the resistance is in a circuit or where the driving V or I comes from. We are not analyzing a whole circuit, just the relationship between V, I, and R for one component.

    P=V*I is valid always. Where V is the voltage difference between two points, and I is the current between the same two points. Between the two points could be an R or a Z or any nonlinear 2-port passive device. For a nonlinear device (think of a Zener diode for example), Ohm's Law may not apply, but P=V*I is valid always.
     
  13. Nov 22, 2014 #12

    phinds

    User Avatar
    Gold Member
    2016 Award

    Yeah, what he said !
     
  14. Nov 22, 2014 #13

    meBigGuy

    User Avatar
    Gold Member

    You are confusing things that are true, and things you have arbitrarily decided to hold constant (like the power)

    Since P = IE, if you increase the voltage, the power increases.
    If you arbitrarily decide the power is to remain constant, then when you increase the voltage. you must also decrease the current for the power to remain constant. Nothing is implied about how that might be accomplished.

    So, E = IR and P=IE and all their combinations are always true.

    As for the physical significance of P=IE, think of it this way.

    When you increase the voltage, the power increases. When you increase the current, the power increases.
    If you have a constant resistance, such as a heating element, if you increase the voltage, the power increases (it get hotter).
    But, note that when you increase the voltage on a constant resistor, the current also increases (since I = E/R) so the power goes up more than you might intuitively expect, as expressed by P = (E^2)/R.

    Think of a room heater with 500W and 1000W settings. Since the effective voltage is constant (AC mains), somehow the current must be varied to achieve different power settings. That can be done by changing the effective resistance (which changes the current). There are lots of ways to do that.

    Hope that helps.
     
  15. Nov 22, 2014 #14
    Things are clear now,thanx for help guys..
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Confusing relation between power,volatage and current
  1. Confusion in current (Replies: 1)

Loading...