1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: A Bit of Help With Peak Current Flowing with Inductor Please

  1. Feb 6, 2010 #1
    Hi can somebody help me with part B of this question please? :)

    2vn3mtx.jpg

    I already did part a) and I found the peak current to be 31.1 mA, and the power rating required to be 4.84 W.

    For part B I know the current for inductance is I=(1/L)[tex]\int[/tex]Vdt
    and I know V=V0sin(wt), however with only this information I don't know what to do, am I suppose to assume the voltage from the previous question and use V0 as 110[tex]\sqrt{}2[/tex]? and am I suppose to assume a frequency of 60 hz (the frequency in the US as stated by my text book), or is there another way to do it.

    And once I find the current, I just use P=IV to find the power right?

    Thanks for any help :D
     
  2. jcsd
  3. Feb 6, 2010 #2

    rl.bhat

    User Avatar
    Homework Helper

    In the pure inductor, the peak current is the same as in the first case. But the power dissipation is zero, because in that circuit the current and the voltage are 90 degrees out of phase and power dissipation W = V*I*cos(theta)
     
  4. Feb 6, 2010 #3
    But how is the peak current the same as in the first case if there is no resistor this time? and is this still assuming the voltage from the first case?
    And would you mind explaining how the current and voltage are 90 degrees out of phase please? Is there an equation?
     
  5. Feb 6, 2010 #4

    rl.bhat

    User Avatar
    Homework Helper

    Sorry. I mean the peak voltage is the same. The peak current will be infinity, because the pure inductor has very small resistance.
    You have written
    I = 1/L*intg(Vo*sinwt) = (Vo/L*w)*(-coswt) = (Vo/L*w)*[sin(w-pi/2)t]
     
  6. Feb 6, 2010 #5
    Thanks. But for the power, P=IV, if I is infinity, I don't understand how the power is zero.
     
  7. Feb 6, 2010 #6
    Is it because when I equals infinity V=0 because they are out of phase? I am confused, how do you know they are out of phase by 90 degrees?
     
  8. Feb 6, 2010 #7

    rl.bhat

    User Avatar
    Homework Helper

    Power is dissipated in the resistance only. If the resistance is nearly zero, power dissipation is also zero.
     
  9. Feb 6, 2010 #8
    OK thanks!
     
  10. Feb 6, 2010 #9
    Can you explain why the voltage and current are 90 degrees out of phase please?
     
  11. Feb 6, 2010 #10

    rl.bhat

    User Avatar
    Homework Helper

    When you connect an inductor to a voltage source, current does not increases suddenly like in resistance, due to back emf. Time taken to reach the maximum current depends on the resistance present in the circuit.
    When the inductor is connected to the ac source the current in the circuit is given by
    I = 1/L*intg(Vo*sinwt) = (Vo/L*w)*(-coswt) = (Vo/L*w)*[sin(wt-pi/2)]
    Or I = Io*[sin(wt-pi/2)] In the resistance I = Io*sin(wt)
    Hence V and I are in phase in the resistance and 90 degree phase difference in inductor.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook