In series with the battery, switch and lamp are two transmission lines. At ##t = 0^+##, we will have twice the characteristic impedance ##2Z_0## of the line in series with the lamp. Since the wire spacing is 1 meter, ##2Z_0## would be quite large, which will limit the current and won't let the lamp glow. Once the voltage steps make one round trip (simultaneously) though the left and right lines, the series impedance will drop to zero and the lamp will light up.
But if we replace the lamp with a neon lamp or an LED, and use a high enough voltage, then the lamp would light up instantly because then the ##2Z_0## would be less than the lamp resistance.
Coupled lines, I've seen myself. I'd crimped a DB25 onto a 300 foot RS232 cable - shielded 6 strand twisted pair. Keyboard input transmitted out the transmit lead (pin 2) would result in matching characters being received as terminal output (pin 3) despite there being nothing plugged in at the far end and no physical connection between the strands.
But if we replace the lamp with a neon lamp or an LED, and use a high enough voltage, then the lamp would light up instantly because then the 2Z0 would be less than the lamp resistance.
Wouldn't that mean that you can send a signal faster than the speed of light?
Wouldn't that mean that you can send a signal faster than the speed of light?
One assumes that a few nanoseconds to get a field across the one meter air gap counts as "instant".
I'm not smart enough to know whether this thing works like a capacitor or a transformer, but it is clear that the connected ends of the wire 1/2 light year away are irrelevant in the near term.
The question is a mares nest, made up of unspecified parameters;
1. The voltage of the battery. It may be 1 V or 1 kV.
2. The impedance of the transmission line. (What diameter are the wires?) 1 m = short circuit.
3. The DC resistance of the very long transmission lines.
4. The cold resistance of bulb B.
5. What is actually meant by “will light up”, a momentary flash, or a continuous dull red glow.
You can get any answer you want by setting different parameters.
Doesn't the current flow generally at something short of the speed of light in vacuum?
That sub-luminal velocity would be along the transmission line if there was a dielectric insulation rather than only vacuum. vf = √ Er .
The near end of the line will appear to be a small capacitance with ½ m long leads, in parallel with the characteristic impedance of the line. Closing the switch might light the lamp for a couple of nanoseconds.
That sub-luminal velocity would be along the transmission line if there was a dielectric insulation rather than only vacuum. vf = √ Er .
The near end of the line will appear to be a small capacitance with ½ m long leads, in parallel with the characteristic impedance of the line. Closing the switch might light the lamp for a couple of nanoseconds.
Forgive my ignorance of EE, but isn't that a basic DC circuit diagram?
The turn on transient is clearly an AC signal.
How long ago was the electrical equipment manufactured and turned on for the first time ?
DC is just very low frequency AC.
The turn on transient is clearly an AC signal.
How long ago was the electrical equipment manufactured and turned on for the first time ?
DC is just very low frequency AC.
If you wait for a year, the reflection from the far end of the two short-circuited transmission lines will get back, then the status of the filament in "bulb B" will be illuminating.
Typical EM propagation speeds in wires is of the order 0.7-0.8 c.
Forget electricity and circuits. Basic relativity tells us that we can't propagate information faster than light. So you can't have a switch that triggers a bulb 0.5 light years away with less than 0.5 years delay. Drawing circuitry in this question is a diversion. The mechanism is irrelevant. Signals can't travel faster than light.
You can get any answer you want by setting different parameters.
This - so not a very interesting question.
There is obviously some capacitance > 0 between the wires, and the more ideal we make all the components the more likely it is that enough current flows for long enough to light the bulb.
Here is a simulation of lines matched to the global load. For time, treat the seconds as years.
The two 1G0 resistors are needed to satisfy the initial DC bias of the transmission line model.
When V1 first rises to 100 V, the 100 ohm globe receives half current because the circuit resistance is 50+100+50=200 ohms. When the reflection of the short circuit finally gets back, the resistance falls to 0+100+0=100 ohms, so the full current flows from then on.
Lowering the line impedance will brighten the globe initially, but reflections will go on forever.
How long ago was the electrical equipment manufactured and turned on for the first time ?
I’m wondering about this too. How is the switch relevant? What I mean is: Does the pulse of current that happens when the switch is closed depend on how long the battery has been hooked up? That is, when you set up this circuit (with the switch open), does the fact that it takes a non-negligible amount of time for the fields to equilibrate globally affect the behavior of the circuit when the switch closes? Or as @jbriggs444 noted, does the local coupling between the wires settle into a steady state relatively quickly?
Another question: does the position of the switch in the circuit matter? If the switch is right beside the battery (and presumably ~1m from the light bulb), then closing it would obviously have a different effect than if the switch was 0.5 light years away.
The state, and the position of the switch in the circuit is relevant because it controls the initial distribution of the electric field that originates at the battery. When the switch is closed the electric field must re-locate, so capacitance must be discharged here while being charged there, with the inductance of the circuit in between limiting the rate of propagation.
In series with the battery, switch and lamp are two transmission lines. At ##t = 0^+##, we will have twice the characteristic impedance ##2Z_0## of the line in series with the lamp. Since the wire spacing is 1 meter, ##2Z_0## would be quite large, which will limit the current and won't let the lamp glow. Once the voltage steps make one round trip (simultaneously) though the left and right lines, the series impedance will drop to zero and the lamp will light up.
But if we replace the lamp with a neon lamp or an LED, and use a high enough voltage, then the lamp would light up instantly because then the ##2Z_0## would be less than the lamp resistance.
I spotted this video the other day and did not get further than him asking this question. I will find time for it later in the week, maybe.
I was (will be) wondering how he deals with it because it is a bit of a can of worms, but what I can say is that the best and most plain-language explanation I have ever seen, which also serves to offer an ideal primer on the impedance of pair/coax lines, is on the following website.
I recommend a read of this link before posing too many follow up questions, because to my mind this answers ALL the questions around this sort of scenario, albeit the reader might have to make one more step to address their particular question/problem but it should put anyone into the right thought process to answer these questions for themselves;-
At the end of the day, if you want to get into bed with electronics at a deeper level of understanding, you should work towards understanding that there is actually no such thing as 'DC'. There is 'quasi-DC' at short timescales and there is what looks like DC on longer timescales when all reactances have found an equilibrium and all resonances damped, and it makes perfect sense to talk in terms of DC for basic real-world electronics at slow switching speeds.
But once you get into 100's kHz then you need to start understanding more stuff on top of what you know about DC. And what you know about DC circuits is virtually useless to any electrical engineering over 2MHz switching rates.
