If DC current is chopped on it's way to a load, what happens to the EM?

AI Thread Summary
Chopping DC power before it reaches a load creates a pulse that travels along the wire, but this action effectively converts the DC signal into an AC-like waveform with a broad frequency spectrum. When the circuit is interrupted, the energy reflects back towards the source, dissipating as heat due to resistance in the wires and load. The electromagnetic (EM) field travels near the speed of light, while electron drift speed remains slow, making momentum a negligible factor in this context. Additionally, the presence of a magnetic field is noted when DC current flows, which changes upon circuit interruption. Understanding the implications of chopping DC is crucial for managing potential interference in electrical systems.
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I am curious what happens if you draw DC power to a load, but as it approaches the load you chop the load connection say with a SSR or switch.

The power will be traveling near the speed of light, following the wire toward the load. When the load is chopped by a switch just before the pulse reaches it, will the power just keep traveling forward, according to Newton's Law of momentum (third law?). Or what comes of it?

I am trying to understand EM in various configurations. I think DC EM may be described as scalar electricity in motion, so cannot be detected by normal means (when measured on it's own). I am attempting to detect it as pulses by various effects on systems. I won't go into it anymore than that for brevity, but thought I'd add some context.

Please let me know if you know the answer. Thanks
 
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Do you mean you switch on the power supply , then before the pulse reaches the load you open another switch just before the load?

In this case, assume the supply has a voltage V. When you switch on, a pulse of amplitude V (a step function) travels from the switch in both directions. The wires constitute a transmission line having a certain characteristic impedance (or surge impedance). The pulse travels in one direction towards the load and in the other direction via the supply to the load. The two pulses are opposite polarity but both end up traveling towards the load on the two wires.

The second switch is open, so the energy on one wire ("live" maybe) is reflected back to the first switch, The energy on the second wire ("ground" maybe) passes through the load and arrives at the opposite side of the second switch where it is also reflected. The energy then travels back to the first switch, and so on for ever. But each time the wave travels over the path it dissipates energy as heat, in the wire resistance, the source resistance and th load resistance, so it soon dies out and we settle to zero voltage on the wires.
 
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username001 said:
The power will be traveling near the speed of light, following the wire toward the load. When the load is chopped by a switch just before the pulse reaches it, will the power just keep traveling forward, according to Newton's Law of momentum (third law?).
Electron drift speed is very slow. The EM travels near light speed, but not the electrons. Momentum plays an insignificant role. So do not attempt to understand electricity as an analogy to bowling balls in a pipe.
 
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username001 said:
The power will be traveling near the speed of light, following the wire toward the load. When the load is chopped by a switch just before the pulse reaches it, will the power just keep traveling forward, according to Newton's Law of momentum (third law?). Or what comes of it?

no it doesnt
There is a pulse of an electric field that travels along the outside of the conductor towards the load
username001 said:
I am trying to understand EM in various configurations. I think DC EM may be described as scalar electricity in motion, so cannot be detected by normal means (when measured on it's own).

DC doesn't produce EM as you suggest it, that requires AC

anorlunda said:
Electron drift speed is very slow. The EM travels near light speed, but not the electrons. Momentum plays an insignificant role. So do not attempt to understand electricity as an analogy to bowling balls in a pipe.

indeed :smile:
 
username001 said:
Please let me know if you know the answer. Thanks
Welcome to PF.
You need to better specify the situation. Where and what do you mean by "chop".
Drag and drop a circuit diagram onto your next post.

A magnetic field will be present if a DC current is flowing. The magnetic field will change when you stop the current by breaking the circuit.
If you are interested in the EM fields of electric power distribution, then you need to study transmission lines.
 
username001 said:
what happens if you draw DC power to a load, but as it approaches the load you chop
One does not simply chop DC :wink:

When you 'chop' it'll make it AC, with a quite wide frequency spectrum. The better your 'chop' is the wider that spectrum, actually. It might result in headaches with antennas, RLC circuits, unbalanced feed lines and such.
 
Been there, done that: Wiring acts as antenna to 'spam' area at chopping-frequency and umpteen multiples...

This is why 'brush' motors must be very carefully suppressed and shielded...
 
Nik_2213 said:
This is why 'brush' motors must be very carefully suppressed and shielded...
... and why the brush covers more than one commutator bar at the time.
 
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