How do we explain powering a load without converting any current?

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Mark Flint

Joined Jun 11, 2017
136
You diagram shows a 24V battery being opposed by a 12V battery so there is 12V left which appears across the bulb.
Why do you think there is none?
It appears we have a voltage difference that is fully explained by the two voltage sources opposing. If we insert an open switch before or after the bulb we have 12v difference across the bulb. If we close the switch we still have the 12v difference across the bulb. Nothing changed because of the bulb. How can we call "no change" a drop?
 

Irving

Joined Jan 30, 2016
3,188
It appears we have a voltage difference that is fully explained by the two voltage sources opposing. If we insert an open switch before or after the bulb we have 12v difference across the bulb. If we close the switch we still have the 12v difference across the bulb. Nothing changed because of the bulb. How can we call "no change" a drop?
No, when you have an open switch you have 12v across the switch and no PD across the bulb as there is no current flow.
 

Thread Starter

Mark Flint

Joined Jun 11, 2017
136
More confusingly, though there will still be approximately 12v across the bulb albeit reversed from the original scenario, there may well be far less current flowing than previously, dependant on the 12v battery's technology and state of charge.
Let's just say standard lead-acid batteries that are 50% charged at 12.4v. In this circuit if you measure a current of 1 amp between the 24v+ terminal and the bulb you will also measure 1 amp between the bulb and the 12v+ terminal.
 

Irving

Joined Jan 30, 2016
3,188
Let's just say standard lead-acid batteries that are 50% charged at 12.4v. In this circuit if you measure a current of 1 amp between the 24v+ terminal and the bulb you will also measure 1 amp between the bulb and the 12v+ terminal.
The actual current is indeterminate - as depends on physical/chemical properties we dont know - but yes, it will be the same wherever you measure it.
 

Irving

Joined Jan 30, 2016
3,188
What I meant to say is that the bulb does not introduce a "drop".
The bulb always introduces a drop of PD = I * R, if I = 0, PD = 0.

If you are conducting a thought experiment and assuming the batteries are ideal, i.e no internal resistance & no chemistry, then PD across bulb with switch closed will be 12v and with switch open, will be zero. But if you are talking real-world then you can't say anything about the actual PD when switch closed as you don't have enough information.
 

Ya’akov

Joined Jan 27, 2019
6,856
There seems to be a hidden assumption that something mysterious is “consumed” by the load. The energy transferred to the load is that stored in the battery’s chemical state

it is transferred to the load via a system of electrical and magnetic fields interacting with electrons in the conductor. Nothing in that transfer involves conversion or consumption.

if the battery was charged by burning something, that’s the conversion/consumption, and it is stored as the chemical state of the battery. When it is discharged, the state returns to its lowest energy.

If we lift a rock and drop it, the kinetic energy is transferred to the rock via the gravitational field. Nothing is converted or consumed to make that happen. The consumption of fuel happened when we lifted the rock, not when we released it.
 

Thread Starter

Mark Flint

Joined Jun 11, 2017
136
There seems to be a hidden assumption that something mysterious is “consumed” by the load. The energy transferred to the load is that stored in the battery’s chemical state

it is transferred to the load via a system of electrical and magnetic fields interacting with electrons in the conductor. Nothing in that transfer involves conversion or consumption.

if the battery was charged by burning something, that’s the conversion/consumption, and it is stored as the chemical state of the battery. When it is discharged, the state returns to its lowest energy.

If we lift a rock and drop it, the kinetic energy is transferred to the rock via the gravitational field. Nothing is converted or consumed to make that happen. The consumption of fuel happened when we lifted the rock, not when we released it.
I have no hidden assumption of something mysterious being consumed by the load. I have no problem with anything in your explanation. My interest is in how energy is release in the load. If there is a "mystery" then this is it. I have heard it said that the voltage drop across the load is responsible for facilitating the energy release to the load. Working with the circuit with two voltage sources it looks like there is no measurable voltage "drop" - only the pre-existing difference in potential between the two voltage sources. So I want to rule in, or out, the premise that energy release is facilitated by a voltage drop across the load itself. May be it's too hard for us lesser mortals to grasp and can only be tackled by the likes of Feynman and theoretical physicists. But that doesn't sit well with me either.
 

Ya’akov

Joined Jan 27, 2019
6,856
I have no hidden assumption of something mysterious being consumed by the load. I have no problem with anything in your explanation. My interest is in how energy is release in the load. If there is a "mystery" then this is it. I have heard it said that the voltage drop across the load is responsible for facilitating the energy release to the load. Working with the circuit with two voltage sources it looks like there is no measurable voltage "drop" - only the pre-existing difference in potential between the two voltage sources. So I want to rule in, or out, the premise that energy release is facilitated by a voltage drop across the load itself. May be it's too hard for us lesser mortals to grasp and can only be tackled by the likes of Feynman and theoretical physicists. But that doesn't sit well with me either.
The voltage drop is an effect not a cause. your proposed circuit does not prevent the voltage drop, it just adds an additional voltage which must be accounted for. If you have the additional voltage that "hides" the voltage drop but no load, the voltage drop caused by the load will not be there and the voltage at that point will reflect its absence.

Cause and effect here are being confused. The voltage drop is an observable effect of the load's consumption of the battery's chemical energy delivered to it by the combination of the E and B fields and the medium of the conductor's free electrons.
 

MrSalts

Joined Apr 2, 2020
2,599
If you put 10 bulbs (12VDC each) in series and connect to a 120v source. Then measure the voltage drop across each bulb. Do this by holding your (AC volt meter meter at on end (keep it there) and then measure the voltage to the point between all the bulbs (across 1, 2, 3...10) bulbs. If you measure any value, you are measuring a potential difference (voltage difference). Report back with your values.
 

crutschow

Joined Mar 14, 2008
31,114
I can’t see anyone trolling this thread. The whole discussion has been civil and I don’t think there has been a single contentious or inflammatory remark made.
A troll doesn't have to be contentious or inflammatory.
It can be someone who continually asks different variations of the same question, while acting as if he doesn't understand the answers.
There have been a few of those on these sites.
 

Deleted member 115935

Joined Dec 31, 1969
0
I have no hidden assumption of something mysterious being consumed by the load. I have no problem with anything in your explanation. My interest is in how energy is release in the load. If there is a "mystery" then this is it. I have heard it said that the voltage drop across the load is responsible for facilitating the energy release to the load. Working with the circuit with two voltage sources it looks like there is no measurable voltage "drop" - only the pre-existing difference in potential between the two voltage sources. So I want to rule in, or out, the premise that energy release is facilitated by a voltage drop across the load itself. May be it's too hard for us lesser mortals to grasp and can only be tackled by the likes of Feynman and theoretical physicists. But that doesn't sit well with me either.
So can we agree that this post is finished with ?

The post is
How do we explain powering a load without converting any current

As we have demonstrated and I think you agree that there is a perfectly logical and well explained reason that current is not "converted" ,
can you close this post

If you have another question, then can I suggest to please start a new thread.
 

Thread Starter

Mark Flint

Joined Jun 11, 2017
136
Hey Eric, Great photo!

I agree. Six pages should have been enough to explain what is going on.
Guys, sure, you can close the post. Thank you to all that have contributed. Even though I still have questions about how energy is delivered this discussion has certainly been beneficial. Cheers.
 
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