Constant Current in DC and AC circuit

[avicenna]

If the voltage across a long wire is constant, is the current uniform throughout the wire length.  

In a 50 Hz ac voltage across a long wire, at a certain moment how does current vary along x, the wire length. 
 

[TheVat]

Is there a load on the circuit, and is it a dumb load (lightbulb, e.g.) or a smart load (electronics)?  When your current passes the load  in a series circuit, there is a voltage drop from hot to neutral (the return leg).  Even without a service load, more than 50 feet of cable (like 12 or 14 AWG in a house) will cause a voltage drop.  Good old R.  (well, Z actually, in an AC circuit)

The voltage drop across the load is proportional to the power available to be converted in that load to some other useful form of energy.

In an AC circuit we speak of impedance.  Several factors at work there.

Besides resistance, AC voltages have a second opposition to current flow called reactance. The sum of resistance and reactance is the impedance.  (Z)  Z will depend on the frequency of the AC and the magnetic permeability of electrical conductors and electrically isolated load elements.  

 

[exchemist]

7 minutes ago, TheVat said:

Is there a load on the circuit, and is it a dumb load (lightbulb, e.g.) or a smart load (electronics)?  When your current passes the load  in a series circuit, there is a voltage drop from hot to neutral (the return leg).  Even without a service load, more than 50 feet of cable (like 12 or 14 AWG in a house) will cause a voltage drop.  Good old R.  (well, Z actually, in an AC circuit)

The voltage drop across the load is proportional to the power available to be converted in that load to some other useful form of energy.

In an AC circuit we speak of impedance.  Several factors at work there.

Besides resistance, AC voltages have a second opposition to current flow called reactance. The sum of resistance and reactance is the impedance.  (Z)  Z will depend on the frequency of the AC and the magnetic permeability of electrical conductors and electrically isolated load elements.  

 

I wonder, though, whether this question may be about something else, viz. the "elasticity" of the current-carrying electrons in the circuit. For instance is the voltage is at its maximum value at one end, what will be the phase of the voltage 10 metres along the wire. Will that also be at the max, or is there a phase lag due to the compressibility of the current carriers? 

[sethoflagos]

52 minutes ago, avicenna said:

If the voltage across a long wire is constant, is the current uniform throughout the wire length. 

I think the correct answer to this rather odd question is yes, though if the wire has any resistance then that uniform current is zero. In the absence of reactive elements (capacitors or inductors) a constant current implies a constant voltage gradient.

52 minutes ago, avicenna said:

In a 50 Hz ac voltage across a long wire, at a certain moment how does current vary along x, the wire length. 

If the load is purely resistive, then the current is in phase with the voltage. A purely capacitive load will make current lead voltage by 90^o. A purely inductive load will cause a 90^o lag. Practical loads fall somewhere in between these extremes.

The variation 'along the wire length' needs to take into account that the electrical field propagates at some substantial fraction of the speed of light, so the wavelength is oto 4,000 km at 50 Hz.

Edited April 24 by sethoflagos
Correction

[TheVat]

I should have mentioned that house loads like tv and computers are more capacitive, and motor stuff (washers, vacs, fridge compressors, induction stoves, ballasts, anything with lots of coils of wire) are inductive loads.  So impedance would depend on which dominates a circuit.  And in houses it is mainly inductive reactance, given the heavy current draw of appliances, furnace blowers, AC, etc.

17 minutes ago, exchemist said:

For instance if the voltage is at its maximum value at one end, what will be the phase of the voltage 10 metres along the wire. Will that also be at the max, or is there a phase lag due to the compressibility of the current carriers?

Good question.  And beyond me, given am not sure how compressibility is defined in this context.  Smaller holes?  🙂

I understand that In AC circuits where there are reactive elements the V and I may not reach the same amplitude peaks at the same time.  This time difference, AKA phase shift, which ranges from 0 to 90° - Seth is saying this isn't significant in the length of a house circuit (usually a single phase circuit, unless your home contains some sort of cottage industry), if I'm following this.  I am mainly (NPI) a guy with some cable and a Klein multimeter who tinkers with house wiring, so I just aspire to be less stupid and not burn the place down.

Edited April 24 by TheVat

[avicenna]

36 minutes ago, sethoflagos said:

I think the correct answer to this rather odd question is yes, though if the wire has any resistance then that uniform current is zero. In the absence of reactive elements (capacitors or inductors) a constant current implies a constant voltage gradient.

I don't  understand. I = V/R, so how can there be zero current if we connect a AA 1.5 V battery to a long copper wire. My question is whether I is the same at all points of the wire.  

[swansont]

1 hour ago, avicenna said:

If the voltage across a long wire is constant, is the current uniform throughout the wire length

If the voltage is constant (for a real wire) there would be no current. If there is a voltage drop, and thus a current, the current will be uniform even if the voltage drop is not (e.g. if there’s a resistor, or a series of different-valued resistors); there’s no way to vary it. Charge is conserved, so current flowing in to a point equals the current flowing out.

[sethoflagos]

2 minutes ago, avicenna said:

I don't  understand. I = V/R, so how can there be zero current if we connect a AA 1.5 V battery to a long copper wire. My question is whether I is the same at all points of the wire.  

V here really stands for voltage difference, not absolute voltage.

So when you say:

1 hour ago, avicenna said:

If the voltage across a long wire is constant ...

Your question reads as if the voltage gradient along the wire is zero. Did you intend that the two ends of the wire are maintained at a constant voltage difference? This would make more sense.

[avicenna]

3 minutes ago, swansont said:

If the voltage is constant (for a real wire) there would be no current. If there is a voltage drop, and thus a current, the current will be uniform even if the voltage drop is not (e.g. if there’s a resistor, or a series of different-valued resistors); there’s no way to vary it. Charge is conserved, so current flowing in to a point equals the current flowing out.

I am saying the potential difference between the ends of the wire, i.e. a long wire connected to a dc battery.

[swansont]

1 minute ago, avicenna said:

I am saying the potential difference between the ends of the wire, i.e. a long wire connected to a dc battery.

The latter part of my explanation applies to this. The current is uniform. There’s no way for it to vary.

[avicenna]

2 minutes ago, swansont said:

The latter part of my explanation applies to this. The current is uniform/\. There’s no way for it to vary.

Thanks. This is what I want to confirm.

For the AC case, I really cannot say as the real world and the ideal world may be different. 

 

[sethoflagos]

1 hour ago, TheVat said:

I understand that In AC circuits where there are reactive elements the V and I may not reach the same amplitude peaks at the same time.  This time difference, AKA phase shift, which ranges from 0 to 90° - Seth is saying this isn't significant in the length of a house circuit (usually a single phase circuit, unless your home contains some sort of cottage industry), if I'm following this.

A domestic dwelling can have an adverse power factor so the phi in pf = cos(phi) is real and measureable especially if it's heavy on the ac and refrigerator loads (as we are, we've got a mechanic tinkering with our knackered gen set as I type!).

The 4,000+ km wavelength I mentioned applies to the time of flight effect over long transmission lines. By the time you see an ac voltage (or current) peak, the last one is long gone. So the phase change over a 10 metre distance is oto 4 second of arc. Good luck measuring that with a multimetre 😊  

[Carrock]

5 hours ago, avicenna said:

If the voltage across a long wire is constant, is the current uniform throughout the wire length.  

In a 50 Hz ac voltage across a long wire, at a certain moment how does current vary along x, the wire length. 
 

There's been a bit of conflation in some replies between A.C. and D.C..  D.C. is conserved but A.C. isn't.

So: V.H.F. A.C. whose effects are more obvious than 50Hz.

You could have 100 meters of 50 ohm coaxial cable with 10db loss per 100 meters. Connecting A.C. power at say 100MHz 100V to the (resistive) cable gives I =V/R = 2 amps input (200W). Connect a 50 ohm resistive load to the far end and you'll get ~ 31.6V at 0.632A i.e. 20W output. If you use a 200 meter cable you'll get 10V at 0.2A i.e. 2W. The input is still 200W into 50 ohms.

The current drops exponentially along the cable. The main power losses are I^R losses in the conductor, dielectric (insulator) heating and radiation from the cable. There is no A.C. current conservation; some of it charges and discharges the dielectric and current is also involved in creating magnetic and electromagnetic fields.

One way of dealing with reactance is to consider the effect of a load impedance mismatch.

e.g. terminate the cable with 25 instead of 50 ohms. This will cause a power reflection back into the cable to compensate for trying to connect a 50 ohm cable to a 25 ohm load. If 20W output then 20 *(50 -25)/(50+25)W i.e. 6.7W is reflected back into the cable and after attenuation 0.67W reaches the source.

You'll get standing waves on the cable; every half wavelength (About 1.5m) you'll get maximum voltage and minimum current; between these nodes you get a minimum voltage, maximum current node. A.C. current can be created and destroyed without breaking conservation laws. Some energy is stored in various fields and doesn't reach the load; sometimes it's called imaginary power(it can be treated as 90 deg or sqrt(-1) out of phase with 'real' power) or reactive power(capacitors and inductors have reactance).

This is sort of real; there are meters which measure forward and reflected power in coax cables...

I didn't want to oversimplify too much; this post ended up much longer than I intended.

[avicenna]

15 hours ago, Carrock said:

There's been a bit of conflation in some replies between A.C. and D.C..  D.C. is conserved but A.C. isn't.

So: V.H.F. A.C. whose effects are more obvious than 50Hz.

You could have 100 meters of 50 ohm coaxial cable with 10db loss per 100 meters. Connecting A.C. power at say 100MHz 100V to the (resistive) cable gives I =V/R = 2 amps input (200W). Connect a 50 ohm resistive load to the far end and you'll get ~ 31.6V at 0.632A i.e. 20W output. If you use a 200 meter cable you'll get 10V at 0.2A i.e. 2W. The input is still 200W into 50 ohms.

The current drops exponentially along the cable. The main power losses are I^R losses in the conductor, dielectric (insulator) heating and radiation from the cable. There is no A.C. current conservation; some of it charges and discharges the dielectric and current is also involved in creating magnetic and electromagnetic fields.

One way of dealing with reactance is to consider the effect of a load impedance mismatch.

e.g. terminate the cable with 25 instead of 50 ohms. This will cause a power reflection back into the cable to compensate for trying to connect a 50 ohm cable to a 25 ohm load. If 20W output then 20 *(50 -25)/(50+25)W i.e. 6.7W is reflected back into the cable and after attenuation 0.67W reaches the source.

You'll get standing waves on the cable; every half wavelength (About 1.5m) you'll get maximum voltage and minimum current; between these nodes you get a minimum voltage, maximum current node. A.C. current can be created and destroyed without breaking conservation laws. Some energy is stored in various fields and doesn't reach the load; sometimes it's called imaginary power(it can be treated as 90 deg or sqrt(-1) out of phase with 'real' power) or reactive power(capacitors and inductors have reactance).

This is sort of real; there are meters which measure forward and reflected power in coax cables...

I didn't want to oversimplify too much; this post ended up much longer than I intended.

I believe this is the real deal, but only for those with some foundations in AC circuit theory. I will see how things relate to what I am investigating.  

@Carrock. 

I am trying to reduce AC circuit to DC circuit "instantaneously". 

I don't want coaxial cables which complicates things. Say I have a simple ac generator that generates fairly good sinusoidal voltage source at 50 Hz (If possible at all?). I connect a long resistive wire to the terminals in a huge circular loop. When the wire is at thermal equilibrium with the environment, we know that all power will be dissipated as IR radiation loss, purely resistive loss - assume ideally.

So we could always apply ohm's law of I=V/R where R is the resistance of the wire, V the instantaneous voltage. It seems that there will be the usual charge conservation along the wire as if the current is a dc current. The current should be a constant at that moment of consideration.

My setup would eliminate capacitance, inductance etc. Instantaneously, we only have the magnetic fields around the wire which we assume "steady". How is such an analysis.
 

[Carrock]

On 4/25/2024 at 2:46 PM, avicenna said:

I don't want coaxial cables which complicates things. Say I have a simple ac generator that generates fairly good sinusoidal voltage source at 50 Hz (If possible at all?). I connect a long resistive wire to the terminals in a huge circular loop. When the wire is at thermal equilibrium with the environment, we know that all power will be dissipated as IR radiation loss, purely resistive loss - assume ideally.

As before, I''ll use values which produce easily described effects.

e.g. 50 Hz A.C. to a 1kW convection heater - the wire to the heater, as well as the heater, will always be warmer than the environment. In this example,  most power will be dissipated as IR radiation but a small amount will be 50Hz radiation from the wiring etc.

On 4/25/2024 at 2:46 PM, avicenna said:

So we could always apply ohm's law of I=V/R where R is the resistance of the wire, V the instantaneous voltage. It seems that there will be the usual charge conservation along the wire as if the current is a dc current. The current should be a constant at that moment of consideration.

My setup would eliminate capacitance, inductance etc. Instantaneously, we only have the magnetic fields around the wire which we assume "steady". How is such an analysis.

dc current is in practice current which has been constant long enough for transients associated with such as inductive or capacitive reactance to become negligible.

'Instantaneous' is not a meaningful concept for measuring e.g. current (coulombs per second) or power (joules per second) - e.g. 0 coulombs in 0 seconds could be any current i.e. not defined.

Calculus, which involves indefinitely small nonzero quantities, is used rather than 'instantaneous' but if applied, everything including inductive and capacitive reactance has to be included.

In particular, alternating current in the wire, and magnetic fields will not be calculated as constant even using indefinitely small nonzero quantities.

 

You might better consider dc or ac separately rather than trying to consider ac as 'almost dc.'

Or perhaps you just want to consider A.C. circuits where reactance and radiative loss are negligible.

[avicenna]

@Carrock, I think your analysis is acceptable. Thanks.