How Can a 100 Ampere Wire Safely Handle a 0.5 Ampere Load?

[kiki_danc]

If you have wire that has 100 ampere current but your load is only 0.5 ampere.. how come the load won't be destroyed by the powerful 100 ampere source?

Or let's take the case of hose. If you have water running so fast inside a 10 inches diameter hose.. and you put a small hose between it.. the small hose can get overwhelmed and break.

Why doesn't the same thing happen to the wire current. What Newtonian physics is involved?

 

[willem2]

Current ratings tend to be maximum ratings. How much current will actually flow depends on the voltage of the source and the resistance of the load.

 

[kiki_danc]

Current ratings tend to be maximum ratings. How much current will actually flow depends on the voltage of the source and the resistance of the load.

If resistance is so low.. why doesn't the current overwhelm it? For example when rebel resistances become so low.. government forces can easily overwhelm it.. how come the same thing doesn't happen in wires?

 

[Nugatory]

If you have wire that has 100 ampere current but your load is only 0.5 ampere.. how come the load won't be destroyed by the powerful 100 ampere source?

If you have 100 amps flowing through the wire to the load, and you only have one load in the circuit, then you have to have 100 amps flowing through the load. This is is because at every point in the circuit, the current in has to be equal to the current.

Thus, the current in the wire and the load will be determined by the voltage of the power source and the resistance of the load, related by Ohm's law: $V=IR$. The way you destroy a device is by applying excessive voltage to it, thereby increasing $I$.

 

[kiki_danc]

If you have 100 amps flowing through the wire to the load, and you only have one load in the circuit, then you have to have 100 amps flowing through the load. This is is because at every point in the circuit, the current in has to be equal to the current.

My home has 100 ampere main breaker.. in the middle of the night.. all gadgets or lights is off.. only my refrigerator is on.. so how come the 100 ampere doesn't overwhelm and destroy the refrigerator?

Thus, the current in the wire and the load will be determined by the voltage of the power source and the resistance of the load, related by Ohm's law: $V=IR$. The way you destroy a device is by applying excessive voltage to it, thereby increasing $I$.

 

[davenn]

My home has 100 ampere main breaker.. in the middle of the night.. all gadgets or lights is off.. only my refrigerator is on.. so how come the 100 ampere doesn't overwhelm and destroy the refrigerator?

your misunderstanding on current is a very common oneThe amount of current flowing in a circuit IS DEPENDANT on the resistance of the circuit, including the load
It IS NOT DEPENDANT on the amount of current the power supply is capable of supplying

for small short distance circuits eg within a device, the resistance of the wires and the circuit tracks can

10V supply and a 10 Ohm load

I (current) = V ( voltage) / R (Resistance)

I = 10 / 10
I = 1 Amp

it doesn't matter if the supply is capable of 10 Amps or 100 Amps ! …. ONLY 1 Amp will flow through the circuit

so to answer your question specifically and not knowing your mains voltage or the resistance (impedance) of the refrigerator
or it's wattage rating I will make some assumptions

240V VAC supply and fridge uses 100 Watts of power

I = W / V
I = 100W / 240V
I = 0.47 A

At this time I have kept it simple and ignored some of the complications of AC voltage
but for basic understanding it's a close enough answer

So although the mains wiring into your house and the main circuit breaker can handle 100 A
There will never be 100A flowing unless you plug in lots of loads ( devices) that all their current consumptions add up to 100 A

Any questions ?

Dave

 

[Nugatory]

My home has 100 ampere main breaker.. in the middle of the night.. all gadgets or lights is off.. only my refrigerator is on.. so how come the 100 ampere doesn't overwhelm and destroy the refrigerator?

A 100 ampere breaker doesn't mean that 100 amps is flowing. It means that if more than 100 amps were flowing the breaker would open the circuit.

The amount of current that is flowing is determined by the resistance of the refrigerator because that's the only thing that's on. The voltage at your main service panel is fixed, the resistance (strictly speaking, impedance because we're talking AC here) of the refrigerator is whatever it is, plug this into Ohm's law and you'll get the amount of current that's flowing. Not only is it much less than 100 amps, it's much less than the 15 or 20 amp (assuming North American standard residential wiring here) breaker on the branch circuit that feeds the refrigerator.

But suppose the voltage at your main service panel were to suddenly increase
(maybe a lightning strike nearby, maybe a malfunction in the power company transformer somewhere in your neighborhood, ...). Then the current would be more, maybe enough more to fry the refrigerator.

In practice the power company is really good at maintaining stable voltages even after equipment failures, the grounding systems of residential wiring minimize the effect of nearby lightning strikes, and appliance manufacturers design their products so that they can tolerate occasional voltage spikes. Thus we don't see many fried refrigerators.

 

[kiki_danc]

your misunderstanding on current is a very common oneThe amount of current flowing in a circuit IS DEPENDANT on the resistance of the circuit, including the load
It IS NOT DEPENDANT on the amount of current the power supply is capable of supplying

for small short distance circuits eg within a device, the resistance of the wires and the circuit tracks can

10V supply and a 10 Ohm load

I (current) = V ( voltage) / R (Resistance)

I = 10 / 10
I = 1 Amp

it doesn't matter if the supply is capable of 10 Amps or 100 Amps ! …. ONLY 1 Amp will flow through the circuit

so to answer your question specifically and not knowing your mains voltage or the resistance (impedance) of the refrigerator
or it's wattage rating I will make some assumptions

240V VAC supply and fridge uses 100 Watts of power

I = W / V
I = 100W / 240V
I = 0.47 A

At this time I have kept it simple and ignored some of the complications of AC voltage
but for basic understanding it's a close enough answer

So although the mains wiring into your house and the main circuit breaker can handle 100 A
There will never be 100A flowing unless you plug in lots of loads ( devices) that all their current consumptions add up to 100 A

Any questions ?

Dave

I know the general formula but why does it work? Think of the electrons in the wire as running in marathon or racing with fast cars.. when the highway becomes small halfway, they should all try to fit in and the highway could be saturated with colliding and damaged cars. So if the resistance is low, the electrons should still run and stampede to it.. but why doesn't this happens.. it looks like the electrons can agree among themselves not to over load the circuit? How do the electrons know when not to stampede in the smaller highway halfway it (the load and resistance or impedance if AC)?

Or what is the microphysics of Ohm's law.. is entanglement involved that is why the electrons can talk among themselves?

 

[russ_watters]

I know the general formula but why does it work? Think of the electrons in the wire as running in marathon or racing with fast cars.. when the highway becomes small halfway, they should all try to fit in and the highway could be saturated with colliding and damaged cars. So if the resistance is low, the electrons should still run and stampede to it.. but why doesn't this happens.. it looks like the electrons can agree among themselves not to over load the circuit? How do the electrons know when not to stampede in the smaller highway halfway it (the load and resistance or impedance if AC)?

Or what is the microphysics of Ohm's law.. is entanglement involved that is why the electrons can talk among themselves?

You seem to be ignoring what people are telling you. Let's try some numbers: If your house's supply voltage is 120V and your refrigerator is the only thing on and uses 500 Watts, what is the current through your 100A main circuit breaker? What is the resistance of the refrigerator (assume DC for simplicity)?

Do you know the two equations you need for this...?

There is no mystery here: it really is this simple.

Or let's take the case of hose. If you have water running so fast inside a 10 inches diameter hose.. and you put a small hose between it.. the small hose can get overwhelmed and break.

This isn't true either, and for the same reason. But let's get the electricity nailed-down first...

 

[kiki_danc]

I'd like to understood it in terms of electrons and atoms.. so I guess my thread title should have been "Microphysics of Ohm's Law". So what happens atomically? Perhaps the electrons are not actually racing.. but only if there is potential difference.. so if the potential difference between the load is small, the electrons won't have tendency to run across it. So they are controlled by potential difference. But what if the potential difference is still low but the load become larger.. I=V/R... so I guess current will decrease. Maybe we can say the electrons are well behave and would only flow if resistance are lower..

 

[kiki_danc]

I'd like to understood it in terms of electrons and atoms.. so I guess my thread title should have been "Microphysics of Ohm's Law". So what happens atomically? Perhaps the electrons are not actually racing.. but only if there is potential difference.. so if the potential difference between the load is small, the electrons won't have tendency to run across it. So they are controlled by potential difference. But what if the potential difference is still low but the load become larger.. I=V/R... so I guess current will decrease. Maybe we can say the electrons are well behave and would only flow if resistance are lower..

Oh.. so for low current light bulb.. resistance is actually larger? (can anyone confirm?) I thought it was smaller...

 

[russ_watters]

I'd like to understood it in terms of electrons and atoms..

No. This approach is not correct. You need to understand this on a macro level first (and English word definition level). Everything else you said in your post is wrong, based on your macro level misunderstanding.

 

[russ_watters]

Oh.. so for low current light bulb.. resistance is actually larger? (can anyone confirm?)

Correct.

 

[kiki_danc]

Correct.

I see. I thought for low current light bulb.. resistance was lower.. so I wonder why all those current won't race to it. So resistance bigger make sense.

At least Newtonian physics or logic is still used.

 

[Dale]

But what if the potential difference is still low but the load become larger.. I=V/R... so I guess current will decrease.

Most electrical power sources are voltage sources, so the potential difference is fixed. Typically 120 V or 240 V depending on the country, or some multiple of 1.5 V depending on the number of batteries. So the potential difference does not change regardless of the current or resistance.

Under those conditions $P=IV=V^2/R$. Since V is fixed, a large load corresponds to a small resistance, and the power and current are controlled by the resistance of the load. By KCL the current through the wire is equal to the current through the load.

I'd like to understood it in terms of electrons and atoms..

You need to understand circuit theory before you can understand Maxwell’s equations, and you need to understand Maxwell’s equations before you can understand quantum electrodynamics.

 

[kiki_danc]

Most electrical power sources are voltage sources, so the potential difference is fixed. Typically 120 V or 240 V depending on the country, or some multiple of 1.5 V depending on the number of batteries. So the potential difference does not change regardless of the current or resistance.

Under those conditions $P=IV=V^2/R$. So a large load corresponds to a small resistance, and the power and current are controlled by the resistance of the load.

Yes, this makes sense. I was asking about the highway or hose with fast racing cars or water... if the resistance is low, they can cross it and stampede...

Or in terms of wire.. if you put a very small wire between two big wires with big current, then the wire melts or explode just like the small diameter hose at center of it bursts. Whereas if you use high resistance between the two wires, then it stops the electrons. Just like you put stone barrier in middle of road or block the hose with stone.. So I guess we can say when the resistance is larger such that it resists the electrons flow.. then you limit the current.

 

[russ_watters]

Yes, this makes sense. I was asking about the highway or hose with fast racing cars or water... if the resistance is low, they can cross it and stampede...

Or in terms of wire.. if you put a very small wire between two big wires with big current, then the wire melts or explode just like the small diameter hose at center of it bursts. Whereas if you use high resistance between the two wires, then it stops the electrons. Just like you put stone barrier in middle of road or block the hose with stone.. So I guess we can say when the resistance is larger such that it resists the electrons flow.. then you limit the current.

This is still wrong. You are not applying what you have learned...

Edit. Ehhhh, it isn't quite the same issue. Your house has different sized wires in it. Why don't the small ones burn up?

Hint: in basic circuit theory, what is the resistance of a wire?

 

[kiki_danc]

This is still wrong. You are not applying what you have learned.

I was expecting you'd say "correct". Or maybe you misunderstood me.

I = V / R

If resistance is less, current across load is more
If highway resistance is less... cars are more
If hose resistance is less, water flows faster

If resistance is more, current across load is less
If highway resistance is more, cars are less
If hose resistance is more, water flows slower

I don't get it. Why am I wrong?

 

[russ_watters]

I was expecting you'd say "correct". Or maybe you misunderstood me.

I = V / R

If resistance is less, current across load is more
If highway resistance is less... cars are more
If hose resistance is less, water flows faster

If resistance is more, current across load is less
If highway resistance is more, cars are less
If hose resistance is more, water flows slower

I don't get it. Why am I wrong?

You are wrong about why puipes burst or wires do or don't burn up. In basic circuit theory what is the resistance of a wire? When a wire burns up what causes it to heat up? How does this differ from the basic assumption in circuit theory?

 

[kiki_danc]

You are wrong about why puipes burst or wires do or don't burn up. In basic circuit theory what is the resistance of a wire? When a wire burns up what causes it to heat up? How does this differ from the basic assumption in circuit theory?

It's like this. When the load is insulator.. no current flows.. it has maximum resistance.. as you lower the resistance, and current flows.. you convert the current to other forms of energy like light or mechanical rotation... When the resistance become zero.. or when you short the wire, it sparks and explodes.

Also what cause a wire to burns up is when it's so small diameter and not matched to the current. In terms of pipes.. when it is too small or weak, and water pressure too strong, the pipe burst.. or wire burns up.

 

[russ_watters]

It's like this. When the load is insulator.. no current flows.. it has maximum resistance.. as you lower the resistance, and current flows.. you convert the current to other forms of energy like light or mechanical rotation... When the resistance become zero.. or when you short the wire, it sparks and explodes.

Also what cause a wire to burns up is when it's so small diameter and not matched to the current. In terms of pipes.. when it is too small or weak, and water pressure too strong, the pipe burst.. or wire burns up.

You are being so sloppy it is tough to tell what you understand and don't understand. Please answer these questions exactly as I have asked them:
1. Under normal circumstances what is the assumed resistance of a wire?
2. If a wire burns up what does that tell us about its resistance?

 

[kiki_danc]

You are being so sloppy it is tough to tell what you understand and don't understand. Please answer these questions exactly as I have asked them:
1. Under normal circumstances what is the assumed resistance of a wire?

Zero in superconductive and a little in ordinary wire.

2. If a wire burns up what does that tell us about its resistance?

Hmm.. it burns up because currents exceeds it capacity.. the resistance is.. hmm.. does resistance increase or decrease when wire burns up?

 

[russ_watters]

Zero in superconductive and a little in ordinary wire.

Correct. We typically assume zero, but in reality it is a low value just above zero.

Hmm.. it burns up because currents exceeds it capacity.. the resistance is.. hmm.. does resistance increase or decrease when wire burns up?

The resistance doesn't change, it just is what it is. But yes, the wire burns up because the resistance is too high.

So put them together:
-A load dissipates more energy when the resistance is low.
-A wire dissipates more energy when the resistance is high.

 

[jbriggs444]

-A load dissipates more energy when the resistance is low.
-A wire dissipates more energy when the resistance is high.

To emphasize what @russ_watters knows well...

We arrange things so that an approximately fixed potential difference is applied to the load. We do this by tapping into an approximately fixed potential difference fed to us by the power company and use wires that are big enough that their resistance is small compared to that of the load.

This means that an approximately fixed current is applied through the wires. Because we used low resistance wires, the current through the circuit is determined primarily by the potential difference supplied by the power company and the resistance presented by the load. It is not much affected by the small resistance presented by the wires.

 

[jim hardy]

you should spend an hour digesting Lavoisier's Preface to his Introduction to Chemistry..
It is a good introduction to scientific thinking.
It's at
https://web.lemoyne.edu/giunta/lavpref.html

Toward the end he quotes an earlier philosopher

""Instead of applying observation to the things we wished to know, we have chosen rather to imagine them. Advancing from one ill founded supposition to another, we have at last bewildered ourselves amidst a multitude of errors. These errors becoming prejudices, are, of course, adopted as principles, and we thus bewilder ourselves more and more. The method, too, by which we conduct our reasonings is as absurd; we abuse words which we do not understand, and call this the art of reasoning. When matters have been brought this length, when errors have been thus accumulated, there is but one remedy by which order can be restored to the faculty of thinking; this is, to forget all that we have learned, to trace back our ideas to their source, to follow the train in which they rise, and, as my Lord Bacon says, to frame the human understanding anew."

so stop trying to force electricity to fit your wild imaginings.

learn the definitions of
Joule
Coulomb
Volt
Ampere

then realize that the water analogy is mediocre one.

Think of current flow analogous to stuffing peas into a straw like you did as a kid to make a pea shooter.
When the straw is full,
For every pea you push into one end, a pea comes out the other end.
But it's not the same pea.
Further, If you block the far end you cannot push any more peas in the near end.

Wire is the same way.
You push a Coulomb in one end and a Coulomb comes out the other.
But it's not the same Coulomb.
The Coulombs push one another along just like peas in a straw. Their actual progress along the straw is a literal snail's pace.
But the 'push' between adjacent Coulombs propagates at a goodly fraction of c.
So the time between a Coulomb entering one end and an identical one popping out the far end is on the order of a nanosecond per foot
(That's why people mistakenly think electrons "zoom down the wire")...
... anyhow...
If you block the load end you can't push any more Coulombs in from the breaker end. .
The degree to which the load end is blocked determines how many Coulombs/Sec per Volt will flow.
Coulombs/sec is Amps ( did you do your homework above?)
Amps per volt is a measure of ease with which current flows into a load and is called "Admittance"
its inverse, the ratio Volts/Amps,
is resistance Ohms.
Ohms is a measure of blockage, ie resistance to current flow hence its name.
Admittance the inverse of resistance has units mho which is ohms spelled backward,

Do your homework and the mysteries will disappear.

 

[Dale]

@kiki_danc this thread is a perfect example of why I dislike analogies for circuits. I assume that you have never actually used the detailed equations that correctly describe the physics of a bursting pipe. You just have this kind of loose analogy in your head. The physics of bursting pipes are substantially more complicated than Ohm's law. So you are using an analogy that is highly complicated and which you don't actually know on a mathematical level to gain insights about a far simpler system. And the insights that you are gaining from the complicated analogy are completely wrong!

I strongly recommend that you discard all of these mental analogies. You are drawing incorrect conclusions from them. Instead focus only on the simple subject of circuit theory itself. For a resistor $V=IR$ and $P=IV$. So if you have a fixed voltage then $P=V^2/R$ and if you have a fixed current then $P=I^2R$. So, in the typical case where you have a fixed voltage then the overall load is larger if the resistance is small. If you have several loads connected in series, then their total resistance will give the current through the series and the resulting current will pass through each series load. Then the single load in the series with the largest resistance will draw the most power.

 

[jim hardy]

I strongly recommend that you discard all of these mental analogies. You are drawing incorrect conclusions from them.

That's the direction he needs to move.

He needs to learn the meaning of the terms before he can understand the equations they go in.
That's why i gave him 'homework'

and referred him to Lavoisier

The impossibility of separating the nomenclature of a science from the science itself, is owing to this, that every branch of physical science must consist of three things; the series of facts which are the objects of the science, the ideas which represent these facts, and the words by which these ideas are expressed. Like three impressions of the same seal, the word ought to produce the idea, and the idea to be a picture of the fact.

and we electrical types perhaps did science a disservice by giving our units names of dead physicists rather than words that produce the idea.

We forget from whence we came - before the word Coulomb automatically painted a picture of 6 E18 X charge of an electron.
That's why i will use analogy to help a newbie struggle up the learning curve as he tries to make the quantities interact properly in his mind..

It has to start with vocabulary
and he needs to do his homework.

old jim

 

[jim mcnamara]

Thread closed, we are at at end point. Thanks for posting.