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Find the rms current

  1. Sep 5, 2013 #1
    1. The problem statement, all variables and given/known data
    The series combination of a 1 kΩ resistor and a 2 H inductor must not dissipate more than 250 mW of power at any instant. Assuming a sinusoidal current with ω=500 rad/s, what is the largest rms current that can be tolerated?


    2. Relevant equations
    Uploaded


    3. The attempt at a solution
    Uploaded. I am confused about finding the current? I am not sure what to do from where I am do I multiply my current value I found by V again and set it equal to .250 W and solve for v then plug back in and solve for I. Which I can the plug into the equation for Ieff.
     
  2. jcsd
  3. Sep 5, 2013 #2

    gneill

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    Staff: Mentor

    You don't need to muck about with voltage if you use an appropriate expression for power.
     
  4. Sep 5, 2013 #3
    woops! I forgot to upload
     

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  5. Sep 5, 2013 #4
    Hmm appropriate expression for power do you mean like P=I^2R
     
  6. Sep 5, 2013 #5

    gneill

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    Yup. The resistor is the only component in the circuit that can dissipate power (inductors store and return power, but they don't dissipate it).

    If the question had asked about the instantaneous apparent power supplied by the current source, then you'd have to worry about the phase of the voltage across the components. But they didn't ask that, they asked for dissipated power.

    If the source is an ideal current supply, then for a series circuit you KNOW that that is the current through all devices, and there's no phase difference for that current for the individual components.
     
  7. Sep 5, 2013 #6
    So do You mean (v/1414)^2 (1000) = .250 but that would be finding voltage?
     
  8. Sep 5, 2013 #7
    Is there anything I can do with my Z value that would allow me to find I without voltage?
     
  9. Sep 5, 2013 #8

    gneill

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    You're looking for the maximum I. Find the peak (instantaneous) power dissipated for a given I.

    Hint: For the instantaneous peak you'll want to use a current value that is a peak (not rms).
     
  10. Sep 5, 2013 #9
    Well instantaneous power is i(t)v(t)=P I found V(t)/1414 = I. I am a little lost what can I do here.
     
  11. Sep 5, 2013 #10
    I(t)=(Vm/1414)cos(wt-45)
     
  12. Sep 5, 2013 #11

    gneill

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    What is the component in the circuit that dissipates power?
     
  13. Sep 5, 2013 #12
    the resistor
     
  14. Sep 5, 2013 #13
    Lets keep going the way we are going but in case it gets too late I have uploaded a solution that I fully understand except two parts.
     

    Attached Files:

  15. Sep 5, 2013 #14

    gneill

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    Right. What's the power dissipated by a resistor with a given current I?
     
  16. Sep 5, 2013 #15
    P(t)=I^2R
     
  17. Sep 5, 2013 #16

    gneill

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    So what current I will dissipate the 250 mW?
     
  18. Sep 6, 2013 #17
    .250=I^2(1000) then I=.015811
     
  19. Sep 6, 2013 #18

    gneill

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    Yes, and using appropriate units: I = 15.81 mA (peak). Always include units when you're presenting results!!!

    To what rms value of current does that correspond?
     
  20. Sep 6, 2013 #19
    15.81/sqrt(2)=11.181 mA
     
  21. Sep 6, 2013 #20
    sorry Ieff=11.181 mA
     
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