4 ms·

Think of the first (current increase) as an acceleration, and the second (current) as a velocity.

by ddebernardy 8y ago

jacquesm 8y ago
4A does not seem all that high?
- thomasfedb 8y ago
  Correct, but it is the rate of change that is more crucial. Just like 60 mph isn't hugely fast, but we still quote 0 to 60 times.
  - jacquesm 8y ago
    Ah now I get it. Thank you very much!
  - amelius 8y ago
    So when I turn on the light at home, what is the increase in current typically, order of magnitude?
    - FPGAhacker 8y ago
      in America household voltage for lighting is usually 110volts AC RMS. And you could divide that by the resistance of the bulb to get current. Or take the power rating and divide that by 110. So a 100 watt bulb is close to an amp. As to the rate, you could use an oscilloscope or a high speed camera.
    - thomasfedb 8y ago
      Decent question, I'm not sure how to calculate that and a cursory Google didn't find me an answer. Household power is AC, I'm sure that will have an impact. For example, switching on the circuit when the voltage is instantaneous 0V would surely result in no change in the flow of current. All you'd have in that scenario would be the rate of change of AC, which would be 10A at 50Hz = 2000 A/s in Australia (1Hz = +240 to -240 and back in 1s). I'm sure this is hopelessly simplistic though, nothing happens instantaneously - there will be some period of high-resistance arking, etc.
      - amelius 8y ago
        I think this concept explains it: https://en.wikipedia.org/wiki/Parasitic_element_(electrical_networks) https://en.wikipedia.org/wiki/Parasitic_element_(electrical_... So instead of modeling your home wiring as wires, you should model them as wires plus capacitances, inductances and resistances. The loop from light switch to lamp and back is basically one "big" inductor. The challenge is to find adequate values for these parasitic elements. You are right about AC though, didn't think of that.
- danielvf 8y ago
  Power is Amps x Volts, so I'm guessing the voltages involved are epicly huge. (Even though this paragraph is really talking about the acceleration)
  - jacquesm 8y ago
    Magnetic fields are all about current and rate of change, the voltage isn't nearly as important a factor as long as it generates the required current. So the biggest element in all this is the internal resistance of the supply, hence the use of capacitors to dump all that energy in one go into a very low impedance coil. I mostly missed the microsecond/million factor, that's 10^12 we're talking about.
    - danielvf 8y ago
      Do you know voltage MRI machines work at?
      - jacquesm 8y ago
        MRI machines are expected to continue to work after they are used once :) The idea here is that the field strength is such that if you don't reach it quickly that the machine will self-destruct before you have reached your target field strength. So you have to cheat, taking for granted that the machine will not survive the experiment, hence the ridiculously high rate of current change. Note the the destruction is totally part of the plan and is a major factor in focusing the field even more. Anyway, the typical MRI power supply is pretty beefy, a few KV and 100's of Amps. And that in a way says it all about how impressive those machines are, they get up to serious field strengths without breaking up.

Think of the first (current increase) as an acceleration, and the second (current) as a velocity.

jacquesm 8y ago

4A does not seem all that high?

thomasfedb 8y ago
Correct, but it is the rate of change that is more crucial. Just like 60 mph isn't hugely fast, but we still quote 0 to 60 times.
- jacquesm 8y ago
  Ah now I get it. Thank you very much!
- amelius 8y ago
  So when I turn on the light at home, what is the increase in current typically, order of magnitude?
  - FPGAhacker 8y ago
    in America household voltage for lighting is usually 110volts AC RMS. And you could divide that by the resistance of the bulb to get current. Or take the power rating and divide that by 110. So a 100 watt bulb is close to an amp. As to the rate, you could use an oscilloscope or a high speed camera.
  - thomasfedb 8y ago
    Decent question, I'm not sure how to calculate that and a cursory Google didn't find me an answer. Household power is AC, I'm sure that will have an impact. For example, switching on the circuit when the voltage is instantaneous 0V would surely result in no change in the flow of current. All you'd have in that scenario would be the rate of change of AC, which would be 10A at 50Hz = 2000 A/s in Australia (1Hz = +240 to -240 and back in 1s). I'm sure this is hopelessly simplistic though, nothing happens instantaneously - there will be some period of high-resistance arking, etc.
    - amelius 8y ago
      I think this concept explains it: https://en.wikipedia.org/wiki/Parasitic_element_(electrical_networks) https://en.wikipedia.org/wiki/Parasitic_element_(electrical_... So instead of modeling your home wiring as wires, you should model them as wires plus capacitances, inductances and resistances. The loop from light switch to lamp and back is basically one "big" inductor. The challenge is to find adequate values for these parasitic elements. You are right about AC though, didn't think of that.
danielvf 8y ago
Power is Amps x Volts, so I'm guessing the voltages involved are epicly huge. (Even though this paragraph is really talking about the acceleration)
- jacquesm 8y ago
  Magnetic fields are all about current and rate of change, the voltage isn't nearly as important a factor as long as it generates the required current. So the biggest element in all this is the internal resistance of the supply, hence the use of capacitors to dump all that energy in one go into a very low impedance coil. I mostly missed the microsecond/million factor, that's 10^12 we're talking about.
  - danielvf 8y ago
    Do you know voltage MRI machines work at?
    - jacquesm 8y ago
      MRI machines are expected to continue to work after they are used once :) The idea here is that the field strength is such that if you don't reach it quickly that the machine will self-destruct before you have reached your target field strength. So you have to cheat, taking for granted that the machine will not survive the experiment, hence the ridiculously high rate of current change. Note the the destruction is totally part of the plan and is a major factor in focusing the field even more. Anyway, the typical MRI power supply is pretty beefy, a few KV and 100's of Amps. And that in a way says it all about how impressive those machines are, they get up to serious field strengths without breaking up.