Simon, I'm sorry, I totally missed your post.
Well this isn't my field at all but a little brush up on the theory tells me its open circuit (assuming the initial pulse is the same polarity).
Let me think about this...
The second before the far end realises, and another second for the current to flow back makes some sort of sense.
So would all three ammeters register the same reading of 1ma after 2 seconds has elapsed ?
And the voltage on meter V4. Steady state there is 150 volts there. Would that voltage appear suddenly only after the 2 seconds has elapsed.
V4. Zero volts initially ? across the earthy end of the cable.
So 150 volts after 2 seconds as it becomes steady state. Is that an "all or nothing" or would you see it initially change and then settle if you looked at it on a scope.
I suppose what I'm asking is whether the potential difference (meter V4) reaches 300 volts (before the current makes has made the round trip). I was thinking like this but a bit unsure of the time scale.
So 150 volts after 2 seconds as it becomes steady state. Is that an "all or nothing" or would you see it initially change and then settle if you looked at it on a scope.
I suppose what I'm asking is whether the potential difference (meter V4) reaches 300 volts (before the current makes has made the round trip). I was thinking like this but a bit unsure of the time scale.
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JN or DF will set me straight with my attempt at an answer.
On closing the switch, electrons will be started to be sucked out of the wire by the battery, and the potential differential this creates will propagate with something like .9E through the wire. This will be noted by A2 and V4 after .55 sec. and by A3 after 1.1 sec.
The question of course is what the meters will read. Me no cable guy. But at t=.55 sec V2 and A2 will only start to rise from zero. At that point in time, there will be a potential gradient of 300 V between A1 and A2. It will take another .55 sec for this gradient to pass through A2, so it will take in total 1.1 sec before it measures 150V.
On closing the switch, electrons will be started to be sucked out of the wire by the battery, and the potential differential this creates will propagate with something like .9E through the wire. This will be noted by A2 and V4 after .55 sec. and by A3 after 1.1 sec.
The question of course is what the meters will read. Me no cable guy. But at t=.55 sec V2 and A2 will only start to rise from zero. At that point in time, there will be a potential gradient of 300 V between A1 and A2. It will take another .55 sec for this gradient to pass through A2, so it will take in total 1.1 sec before it measures 150V.
The potential difference propagating down the wire sounds very plausible. I'd add to that and go so far as to say no current would show on any meter during this time frame. This "wavefront" then reaches the midpoint of the cable and V4 registers 300 volts. The wavefront keeps travelling and only when it reaches the battery negative does the current flow. All three ammeters then read 1ma and V4 falls to 150 volts... hmm but thinking about it, it can't fall instantly.
The more you think about it the more questions it raises
No, I don't think there will ever be a 300 V reading on V4, it will stop at 150 V. As the potential gradient passes through A2 it will broaden from A1 to A3 and the midpoint will be 150 V.
On the amps: A1 will go from infinity at switch on to 1 mA in 1.1 sec. At .55 sec. it will be 2 mA.
A1 will be 2mA at t=.55 sec and 1 mA at t=1.1 sec, when A3 will jump from 0 to 1 mA.
Btw. as SY mentioned all this without taking inductance into account.
On the amps: A1 will go from infinity at switch on to 1 mA in 1.1 sec. At .55 sec. it will be 2 mA.
A1 will be 2mA at t=.55 sec and 1 mA at t=1.1 sec, when A3 will jump from 0 to 1 mA.
Btw. as SY mentioned all this without taking inductance into account.
Thomas Townsend Brown - Wikipedia, the free encyclopedia
I've posted it before ......... I'll post it again .....
I've posted it before ......... I'll post it again .....
Something missing from the diagram is capacitance - all along the cable. The battery will also have some capacitance. Capacitance to what? Ground, or the rest of the universe. Capacitors are strange, because they can have just one terminal.Mooly said:
A thought experiment must be physically complete and consistent. Apparent anomalies often arise from forgetting this. So treat the wires as though they are the inner of a coaxial cable, with ground/universe as the outer. Then it becomes a transmission line problem, which can be solved using standard techniques.
Bear in mind that the battery has two terminals so a current will flow in both of them when the switch is closed.
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