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

Work that can be done via induction (on/off) vs. the electrical power used to make it

  1. Feb 11, 2008 #1
    Energy stored in an inductor is equal to:

    [itex]\frac{1}{2}L\mathbf{I}^2[/itex]

    Where [itex]L[/itex] is the inductance in henries and [itex]I[/itex] is the current in amps.

    The energy stored in the inductor doesn't depend on how fast the current is attained, just the fact that it has some inductance and some current.

    However, how fast the current can be attained can very well depend on what is supplying the inductor with current. The timing of turning on and off of the inductor magnetic field affects how much work could done by it. So what if we take the energy of the inductor and multiply by the frequency of the pulses? This should not exceed overall electrical power [itex]R{I}^2[/itex] if current electromagnetic theory is correct.

    Assuming power factor is 1 (or not assuming power factor is one), [itex]\frac{1}{2}Lf[/itex], where [itex]f[/itex] is frequency of on/off periods the inductor handles (to influence a magnetic rotor), cannot be greater than the resistance [itex]R[/itex] of the coil. Do I have this right?

    For example it should be impossible that a coil of 1100 henries with 770 Ohms of resistance can switch its whole magnetic field (i.e. with the current throughout the whole length of the coil) on and off 1.4 times every second. Isn't this well understood in engineering literature (I hope it is)? So [itex]\frac{1}{2}Lf \le R[/itex]?

    The time constant of any motor is simply inductance divided by resistance. This would mean that the time constant of the circuit times the frequency of the pulses cannot be greater than 2. Right?
     
    Last edited: Feb 11, 2008
  2. jcsd
  3. Feb 16, 2008 #2
  4. Feb 19, 2008 #3
    1/2*Inductance*(change of current from zero)^2/(time to change current) = power of an inductor

    1/2*(Inductance/Resistance)*voltage*(change of current from zero)/(time to change current) = power of an inductor

    (1/2*(time constant of circuit)/(time to change current))*voltage*(change of current from zero) = power of an inductor

    power of inductor / electric power sent to coil = 1/2*(time constant of circuit)/(time to change current)
     
    Last edited: Feb 19, 2008
  5. Feb 19, 2008 #4

    russ_watters

    User Avatar

    Staff: Mentor

    It just looks to me like you are mixing unrelated concepts there: Inductive devices don't have a high resistance, it would defeat their purpose. So you don't use ohm's law on them.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Work that can be done via induction (on/off) vs. the electrical power used to make it
Loading...