I'm trying to wrap my head around amps. I get volts and I get the theory of amps. What I'm really lacking is a practical understanding of amps. How do volts relate to amps? I know V X I = Wattage. I don't understand why 12v SLA batteries have such high amps in relation to other 12v circuits. I also don't understand why my coffee maker is 120v but only 7 amps. I must be really missing some key understanding. Thanks.
Could someone please really break down amps for me?
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BTW I'm assuming my coffee maker has a step down transformer in it. It doesn't really run off 120 volts. Right? That would be hard to believe. But what do I know. I've never cracked one open 
You say you get the theory of amps, what is it that you understand about it? To put it simply, amperage (current), is the flow of charge. I think the easiest way to picture it is what happens in a capacitor.
http://www.flashscience.com/electricity/capacitor.htm
Basically, if you look at just the capacitor, when the circuit is shorted so the capacitor is excluded, there is no current or voltage on the capacitor branch. However, once you connect the capacitor, there is a large current (amperage) as electrons flow to the capacitor and build on its plates. As the electrons build, the voltage increases because there is greater potential from the separation of charges. Once the capacitor reaches its charge limit, there is no more current (amperage) because the capacitor's plates are full and can't handle any more electrons, but the voltage is at its maximum because of the build up of charges. If you disconnect the battery from the circuit, the process reverses as current flows to the light build to light it (and some will be lost through the resistor too). At some point if you were to forcefully raise the voltage across the capacitor, there would be a dielectric breakdown and a large current, but that's another topic.
Any questions? Did this help?
http://www.flashscience.com/electricity/capacitor.htm
Basically, if you look at just the capacitor, when the circuit is shorted so the capacitor is excluded, there is no current or voltage on the capacitor branch. However, once you connect the capacitor, there is a large current (amperage) as electrons flow to the capacitor and build on its plates. As the electrons build, the voltage increases because there is greater potential from the separation of charges. Once the capacitor reaches its charge limit, there is no more current (amperage) because the capacitor's plates are full and can't handle any more electrons, but the voltage is at its maximum because of the build up of charges. If you disconnect the battery from the circuit, the process reverses as current flows to the light build to light it (and some will be lost through the resistor too). At some point if you were to forcefully raise the voltage across the capacitor, there would be a dielectric breakdown and a large current, but that's another topic.
Any questions? Did this help?
Put simply, it is the electron flow rate. Just like a big hose might have 1 gallon/s flowing through it, a circuit might have 5 Amps.
Literally: 1 AMP = 6.241 × 10^18 electrons (1 coulomb) flowing per second.
It can be very pressurized (high voltage), or not as pressurized (low voltage). It all depends on the resistance. How hard it is for it to flow is resistance, and the amount of work being expended to overcome that resistance is power.
Literally: 1 AMP = 6.241 × 10^18 electrons (1 coulomb) flowing per second.
It can be very pressurized (high voltage), or not as pressurized (low voltage). It all depends on the resistance. How hard it is for it to flow is resistance, and the amount of work being expended to overcome that resistance is power.
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What I want to understand is how amperage actually affects the physical work in the real world. You explained how amps work in the circuit with the capacitor. I get that amps measure the flow of charge, and volts are the potential difference. What I'm trying to understand is how this actually affects appliances and physical energy.
Let me put it this way: if you had a motor that spins off electricity that was on a 3v 1amp circuit, would it spin faster if you increased the Voltage or increased the amps?
Let me put it this way: if you had a motor that spins off electricity that was on a 3v 1amp circuit, would it spin faster if you increased the Voltage or increased the amps?
Current (measured in amps) can be thought of as the rate at which electrical charge flows through a circuit. So current measured in amps, is analagous to water flowing through a pipe measured in gallons per second. In fact, 1 amp means a flow of 6,241,000,000,000,000,000 electrons per second.
Potential difference (measured in volts) can be thought of as the "pressure" with which electricity tries to flow, given a circuit to flow through. The higher the potential difference, the more "force" gets applied to the electrons as they try to flow through a material. Some materials like metals allow electrons to flow very easily with very little force. Others like graphite or silicon, resist the flow of electrons and electrons tend to flow slowly unless there is a large potential to force them through. Others like ceramics and plastics are almost completely impervious to electrons (these are electrical insulators), so none flow, even in the presence of a high potential difference.
So the current depends fundamentally on two things:
1. The applied potential
2. The resistance of the circuit.
Ohm's law essentially states that current = potential / resistance.
As you can see from that equation. If you increase the potential, the current increases as well (if resistance stays the same). If you increase resistance, the current drops (if potential stays the same).
Energy and power (the rate at which energy is converted) are related to current and potential difference.
Power = current * voltage.
So, a coffee maker which takes 10 A when connected to 120 V - has a power of 1200 Watts.
You also apply Ohm's law and work out that the coffee maker has a resistance of 12 Ohms (120 / 10).
If you connected the coffee maker to a 12 V car battery, then it would take 1 A (12 V/ 12 Ohm). When connected to the car battery the coffee maker would have a power of 12 V * 1 A = 12 W.
Potential difference (measured in volts) can be thought of as the "pressure" with which electricity tries to flow, given a circuit to flow through. The higher the potential difference, the more "force" gets applied to the electrons as they try to flow through a material. Some materials like metals allow electrons to flow very easily with very little force. Others like graphite or silicon, resist the flow of electrons and electrons tend to flow slowly unless there is a large potential to force them through. Others like ceramics and plastics are almost completely impervious to electrons (these are electrical insulators), so none flow, even in the presence of a high potential difference.
So the current depends fundamentally on two things:
1. The applied potential
2. The resistance of the circuit.
Ohm's law essentially states that current = potential / resistance.
As you can see from that equation. If you increase the potential, the current increases as well (if resistance stays the same). If you increase resistance, the current drops (if potential stays the same).
Energy and power (the rate at which energy is converted) are related to current and potential difference.
Power = current * voltage.
So, a coffee maker which takes 10 A when connected to 120 V - has a power of 1200 Watts.
You also apply Ohm's law and work out that the coffee maker has a resistance of 12 Ohms (120 / 10).
If you connected the coffee maker to a 12 V car battery, then it would take 1 A (12 V/ 12 Ohm). When connected to the car battery the coffee maker would have a power of 12 V * 1 A = 12 W.
Note, in Mark's examples above, those equations only apply to purely DC circuits. An AC circuit, most motors, work off of slightly different but more complex equations involving complex numbers. Just to be clear.
The vast majority of power is supplied by voltage sources (I don't know of anything that runs off of a current source but maybe it exists). What this means is the voltage is fixed and since the circuit isn't changing the resistance is also fixed so the 1A is just a result of the circuit's resistance. You can't just increase the amps, they all play together. So with V and R constant V/R=I and you can't just change I. You can change the circuit and reduce R which would increase the current or you can increase the voltage and increase the current.
The confusion comes, I think, when you see a power supply say 12V/1A or something similar. What you have to remember is that it is designed to always provide 12V and can provide up to 1A but if hooked up to something with R=24ohm it will only provide 0.5A for example.
All of this is ignoring AC supplies and I'm skating the motor question as I can't remember anything about how DC motors work off the top of my head.
My thanks to everyone for the answers. Thanks to you guys, this concept is becoming clearer.
My dilemma is that I don't do well at rote learning. I like to learn by doing. Working out equations in my head is a drag. So essentially, the challenge I'm facing is to take what is occurring at the microscopic level, and translate that into the macroscopic.
For example, automobiles are easy to understand because we can see, visually, the mechanical work that is taking place. Then the concept of an automobile becomes concrete and physical, not abstract.
When we look at electricity, it isn't something many people can pick up easily, because you don't see it. It is mostly in the conceptual level. So that is why it's confusing, to me.
At this point, thanks to you guys, what do I understand about the "physics" of electricity? Well, I'm visualizing all materials as made up of small particles called atoms. And "conductors" are materials with atoms that act like bridges, which electrons can cross.
The more bridges you have, ie the more conductive atoms, the more electrons can pass. Right? So, the thicker a wire is, the higher it's amp rating. Because thicker has more bridges available than thinner. Is that about right?
And now I'm seeing lots of other stuff, too. Voltage is the potential difference that "pushes" the free electrons across.
I guess amps really are best understood in context of circuits and Ohm's law. Is that right? Then you can calculate the current flow.
I'm still not getting how amps are manipulates by hobbyists and that's what I want to understand. If you want to change voltages you simply change a battery. To increase resistance you add a resistor. If you want to increase amps, then you would simply use thicker wires? Is that right?
Many thanks guys.
My dilemma is that I don't do well at rote learning. I like to learn by doing. Working out equations in my head is a drag. So essentially, the challenge I'm facing is to take what is occurring at the microscopic level, and translate that into the macroscopic.
For example, automobiles are easy to understand because we can see, visually, the mechanical work that is taking place. Then the concept of an automobile becomes concrete and physical, not abstract.
When we look at electricity, it isn't something many people can pick up easily, because you don't see it. It is mostly in the conceptual level. So that is why it's confusing, to me.
At this point, thanks to you guys, what do I understand about the "physics" of electricity? Well, I'm visualizing all materials as made up of small particles called atoms. And "conductors" are materials with atoms that act like bridges, which electrons can cross.
The more bridges you have, ie the more conductive atoms, the more electrons can pass. Right? So, the thicker a wire is, the higher it's amp rating. Because thicker has more bridges available than thinner. Is that about right?
And now I'm seeing lots of other stuff, too. Voltage is the potential difference that "pushes" the free electrons across.
I guess amps really are best understood in context of circuits and Ohm's law. Is that right? Then you can calculate the current flow.
I'm still not getting how amps are manipulates by hobbyists and that's what I want to understand. If you want to change voltages you simply change a battery. To increase resistance you add a resistor. If you want to increase amps, then you would simply use thicker wires? Is that right?
Many thanks guys.
John Connor
Lifer
- Nov 30, 2012
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Yeah that thought did occur to me immediately after I made my 2nd post. Heating element to boil the water, duh
2timer,
While not really an exactly accurate description, you're definitely on the right track. I won't get into the actual math/physics of electron flow, but that's a good way to look at it from a starting point. Just note, electrons don't need an atom to act like a bridge in order to travel everywhere they go.
As far as current manipulation, I think you have enough to figure it out for a basic circuit
V = I*R (again, DC only)
a) If you have room to increase voltage to achieve a higher current, you can do that, assuming resistance stays the same.
b) If you need a set voltage, then you can decrease resistance to give you increased current. Using a thicker wire would have this effect, though typically, anything substantial would require a very thick wire. Thicker wires are used because they can handle more current without heating up too much (current causes heat according to R * I^2) but this starts to get into the physics of it again. I can explain it though if you want.
Both of the above, however, assume a very, very basic circuit which would do essentially nothing for you in real life. There's actually all sorts of fun games you can play with circuits to manipulate voltage and current. A very basic example is if you put two identical resistors in parallel, both connected to a voltage source. This would split your current between the two resistor branches so that each branch receives half of the original current.
Now, if R2 is say, a light bulb (your load circuit), and before you hook up R1, you calculate that R2 has 0.1 amps of current going through it when connected to an ideal 10 V battery (with no internal resistance). You would then know that your load (light bulb) has a resistance of 100 ohms. Thus, by choosing R1 to be 100 ohms and placing it in parallel with R2 as shown above, you now have 0.1 / 2 = 0.05 amps running through your light bulb. By making R1 bigger or smaller, you can adjust the amount of current going through the light bulb theoretically from 0 amps (R1 is a wire with no resistance) to 0.1 amps (you make R1 infinite).
In electronics, current is pretty much controlled by transistors (BJTs and MOSFETs), but that's a whole different discussion.
While not really an exactly accurate description, you're definitely on the right track. I won't get into the actual math/physics of electron flow, but that's a good way to look at it from a starting point. Just note, electrons don't need an atom to act like a bridge in order to travel everywhere they go.
As far as current manipulation, I think you have enough to figure it out for a basic circuit
V = I*R (again, DC only)
a) If you have room to increase voltage to achieve a higher current, you can do that, assuming resistance stays the same.
b) If you need a set voltage, then you can decrease resistance to give you increased current. Using a thicker wire would have this effect, though typically, anything substantial would require a very thick wire. Thicker wires are used because they can handle more current without heating up too much (current causes heat according to R * I^2) but this starts to get into the physics of it again. I can explain it though if you want.
Both of the above, however, assume a very, very basic circuit which would do essentially nothing for you in real life. There's actually all sorts of fun games you can play with circuits to manipulate voltage and current. A very basic example is if you put two identical resistors in parallel, both connected to a voltage source. This would split your current between the two resistor branches so that each branch receives half of the original current.
Now, if R2 is say, a light bulb (your load circuit), and before you hook up R1, you calculate that R2 has 0.1 amps of current going through it when connected to an ideal 10 V battery (with no internal resistance). You would then know that your load (light bulb) has a resistance of 100 ohms. Thus, by choosing R1 to be 100 ohms and placing it in parallel with R2 as shown above, you now have 0.1 / 2 = 0.05 amps running through your light bulb. By making R1 bigger or smaller, you can adjust the amount of current going through the light bulb theoretically from 0 amps (R1 is a wire with no resistance) to 0.1 amps (you make R1 infinite).
In electronics, current is pretty much controlled by transistors (BJTs and MOSFETs), but that's a whole different discussion.
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Following the same basic example, if you instead put the resistors in series:
Where R1 = R2, the voltage in the wire between R1 and R2 will be half of your voltage source. So, if you want to manipulate a voltage to be between 0 and your original voltage source, you can play with resistors to get any voltage drop you want (theoretically) by
V2 = V * (R2 / (R2 + R1)) where V is your original voltage and V2 is your voltage between R1 and R2.
Oh, I forgot, in the previous post, you get the current by:
I2 = I * (R1 / (R1 + R2)) where I is your originally measured current and I2 is the current going through R2.
Also note, both of the above examples would be a horrible voltage/current source, this is just to get the basics of V = I * R and how you can play with it.
Where R1 = R2, the voltage in the wire between R1 and R2 will be half of your voltage source. So, if you want to manipulate a voltage to be between 0 and your original voltage source, you can play with resistors to get any voltage drop you want (theoretically) by
V2 = V * (R2 / (R2 + R1)) where V is your original voltage and V2 is your voltage between R1 and R2.
Oh, I forgot, in the previous post, you get the current by:
I2 = I * (R1 / (R1 + R2)) where I is your originally measured current and I2 is the current going through R2.
Also note, both of the above examples would be a horrible voltage/current source, this is just to get the basics of V = I * R and how you can play with it.
sm625
Diamond Member
- May 6, 2011
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You have two ways of getting the same amount of power. You can provide a very high voltage and low current, or a very high current and low power. A lead acid battery is an example of low voltage high current. The 1.2V regulators on a motherboard are an even better example. In some cases you can exceed 100 amps while only pushing 1 volt. Despite all those amps, you're only looking at around 100 watts.
On the other end of the spectrum you have something like a taser which delivers high voltage and low current. A taser might run off a 4V battery and put out 40,000 volts at 100 microamps. Likewise, it might pull 1 amp from that 4V battery for a total of 4 watts, but the person being tazed will be getting that 4 watts (minus some losses) in the form of high volts at low current. As you know it only takes a few milliamps to kill. Same goes with ignition coils in a car. High volts, low current. As long as you know the power at any point in the conversion you can calculate both volts and current in a circuit on either side of a coil.
On the other end of the spectrum you have something like a taser which delivers high voltage and low current. A taser might run off a 4V battery and put out 40,000 volts at 100 microamps. Likewise, it might pull 1 amp from that 4V battery for a total of 4 watts, but the person being tazed will be getting that 4 watts (minus some losses) in the form of high volts at low current. As you know it only takes a few milliamps to kill. Same goes with ignition coils in a car. High volts, low current. As long as you know the power at any point in the conversion you can calculate both volts and current in a circuit on either side of a coil.
The lethal amperage window is actually in the hundreds of milliamps (100-300 usually quoted). Below that, you'll have pain but it won't be lethal. Above that, you'll have severe pain/burns but most likely it won't be lethal either though depending on the shock, someone might have to help you start breathing again. Of course, if you get really high amperage, you just fry anyway. The lethal window comes because the current passing through the heart in that window causes heart palpatations and then the heart just locks up because the body can't control its frequency any more, essentially a heart attack.
Hitman928, that was a fabulous explanation, thank you. Not to detract in the slightest from what other people said. Perhaps I just was finally effective in asking the right question?
Now I understand. Current is essentially a pathway across which a charge travels. So in a circuit, the current is calculated by dividing voltage by the amount of resistance... Wow.
I can see how all this can escalate in difficulty quite easily. Clearly, I am a novice in electronics and I have nowhere near the knowledge of others on here
So, to sum everything up, current or amps in a simple DC circuit is just a relative of the main function of voltage and resistance.
This still doesn't answer the original question about the car battery. If amps are a measure of current flow, then how can a car battery be rated in amps? A car battery is just a voltage source, not a circuit.
Now I understand. Current is essentially a pathway across which a charge travels. So in a circuit, the current is calculated by dividing voltage by the amount of resistance... Wow.
I can see how all this can escalate in difficulty quite easily. Clearly, I am a novice in electronics and I have nowhere near the knowledge of others on here
So, to sum everything up, current or amps in a simple DC circuit is just a relative of the main function of voltage and resistance.
This still doesn't answer the original question about the car battery. If amps are a measure of current flow, then how can a car battery be rated in amps? A car battery is just a voltage source, not a circuit.
The amp rating is the max current. It can provide up to X amps before "bad stuff" starts happening (bad stuff could be as benign as it drops the voltage to it catches fire).
It's far more complex than that. A "voltage source" is actually an "electron source" as well. It can only spit out so many electrons at a time, regardless of the resistance.
True but if a battery is rated for 10A that doesn't mean it can't spit out more electrons than 10A it means that if you put a load on it such that V/R > 10 it will no longer function correctly. It might cause too much heat and burn up or it might cause the voltage to droop. You'd have to know about the battery design to know what will happen when you cross the given limit.
Up until you get to that 10A point you can treat it as a voltage source ie it provides X volts.
Murloc
Diamond Member
- Jun 24, 2008
- 5,382
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Physics without maths can't be studied. You can still get the concepts without maths though, understanding stuff like how an ideal battery with internal resistance behaves is beyond that already though, imho.
Want macroscopic stuff?
Try out the hydraulic analogy: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/watcir.html
Want macroscopic stuff?
Try out the hydraulic analogy: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/watcir.html
Ahhhh so it is LOAD that is measured in amps, ie how many columbs of charge are actually making "x" device work? If this simplified explanation is correct, that would make a lot more sense for me, ie, I could view a home appliance as being essentially comprised of: a) a voltage source , b) the circuit components which control the current flow, and c) the output end/load device (eg, the heating element in a coffee maker, the motor in a DC motor circuit, or any number of transducer outputs in DC circuits). I KNOW this is very simple, but it at least is logical, right?
I ALMOST UNDERSTAND MY COFFEE MAKER NOW!!! LOL. My God, trying to understand electricity in terms of physics is exhausting. I almost want to quit...I can't quit...
Must...
go...
on...
I ALMOST UNDERSTAND MY COFFEE MAKER NOW!!! LOL. My God, trying to understand electricity in terms of physics is exhausting. I almost want to quit...I can't quit...
Must...
go...
on...
There are some misleading facts in this thread.
So I will try to correct them.
And it's good to explain first how energy is carried from hard sources (wall receptacles) when you understand this, than you will move on batteries(soft sources)
1. Voltage, denoted oftenly as U is measured in Volts(V) - is the primary force that causes electricity to work, it polarizes electrons in conductors connected to the voltage source and they are ready to flow once the circuit is closed.
2. Current, denoted as I is measured in Amperes(A) - is series of moving free-stream electrons in electrically-conductive materials, when they are exposed to voltage source. Making it carrier of electrical charge, constant flow of electrical charge transfers energy. There are 2 types of current, the Alternating current(AC) and Directive Current(DC). About when and why they are used later. What is important is that current starts to flow only when circuit connected to voltage source is closed.
3. Power(electrical energy), denoted as P, also known as Performance and measured in Watts(W) - is universal unit for describing how much energy is required to run particular device. It is the only important factor we need to overcome challenges we are working on when constructing and deploying electrical appliances. All other units, including voltage, resistance and current are here only to optimize efficient performance of appliances.
And now let's go on how voltage and current work together.
Voltage is force that cause current to flow because it polarizes the electrons inside the conductors, when you close the circuit, they start to flow and start transfering electrical energy to the devices within the closed circuit, creating current, the multiplication of voltage and current will tell you how much energy in watts we are using. The correlation between voltage and current depends on power drawn. For example if your coffee maker has 120V/7A rating on its label it draws 840W from wall receptacle. If you would however connect it to 230V(in case its designed for that) it would draw only 3.6A, but the transfered energy would still be same 840W. So in linear correlation with lower voltage the higher current is needed, if the voltage is higher than current is automatically lower just to hit the required energy for the device.
It depends on the purpose of device how much energy it uses, the devices required for heavy applications need much more power than the household ones.
Above mentioned coffee maker draws 840W which is quite alot of energy, because it needs to boil the water, it must generate alot of heat, as the current that flows through the resistive spiral must be high in order for it to become hot and therefore heat the water for coffee.
The cellphone a small lightweight device, while charging uses no more than 5W of energy.
Light rail vehicle which is quite heavy on its own and travels at high speeds draws as much as 500kW(500 000W)of energy and is by that way also powered under higher voltage(oftenly from 600 to 2500V) and uses current as much as 800A when accelerating and braking.
Electric locomotive uses upto 4000kW, and may be connected to catenary powered by as much as 25kV(25 000V) and utilize current upto 1500 A.
On other hand, the nuclear reactor in power plant has performance of 950MW(950 000 000 W).
What I mentioned now were only maximal values, the energy required changes depending on how much load the device has to carry on.
The uncoupled locomotive will draw only 50kW from catenary, when attached to a long fully loaded freight train it will use its 4000kW like nothing.
The idle computer draws about 50W, when you turn on the games or heavy programs it may go as high as 1kW depends on the parts inside.
Coffee maker or light bulb are passive and they use same amount of energy everytime they are turned on.
If you need to know something about electricity feel free to ask but it's way too much to just write it here all
Or if someone knows better correct me if I'm mistaken.
So I will try to correct them.
And it's good to explain first how energy is carried from hard sources (wall receptacles) when you understand this, than you will move on batteries(soft sources)
1. Voltage, denoted oftenly as U is measured in Volts(V) - is the primary force that causes electricity to work, it polarizes electrons in conductors connected to the voltage source and they are ready to flow once the circuit is closed.
2. Current, denoted as I is measured in Amperes(A) - is series of moving free-stream electrons in electrically-conductive materials, when they are exposed to voltage source. Making it carrier of electrical charge, constant flow of electrical charge transfers energy. There are 2 types of current, the Alternating current(AC) and Directive Current(DC). About when and why they are used later. What is important is that current starts to flow only when circuit connected to voltage source is closed.
3. Power(electrical energy), denoted as P, also known as Performance and measured in Watts(W) - is universal unit for describing how much energy is required to run particular device. It is the only important factor we need to overcome challenges we are working on when constructing and deploying electrical appliances. All other units, including voltage, resistance and current are here only to optimize efficient performance of appliances.
And now let's go on how voltage and current work together.
Voltage is force that cause current to flow because it polarizes the electrons inside the conductors, when you close the circuit, they start to flow and start transfering electrical energy to the devices within the closed circuit, creating current, the multiplication of voltage and current will tell you how much energy in watts we are using. The correlation between voltage and current depends on power drawn. For example if your coffee maker has 120V/7A rating on its label it draws 840W from wall receptacle. If you would however connect it to 230V(in case its designed for that) it would draw only 3.6A, but the transfered energy would still be same 840W. So in linear correlation with lower voltage the higher current is needed, if the voltage is higher than current is automatically lower just to hit the required energy for the device.
It depends on the purpose of device how much energy it uses, the devices required for heavy applications need much more power than the household ones.
Above mentioned coffee maker draws 840W which is quite alot of energy, because it needs to boil the water, it must generate alot of heat, as the current that flows through the resistive spiral must be high in order for it to become hot and therefore heat the water for coffee.
The cellphone a small lightweight device, while charging uses no more than 5W of energy.
Light rail vehicle which is quite heavy on its own and travels at high speeds draws as much as 500kW(500 000W)of energy and is by that way also powered under higher voltage(oftenly from 600 to 2500V) and uses current as much as 800A when accelerating and braking.
Electric locomotive uses upto 4000kW, and may be connected to catenary powered by as much as 25kV(25 000V) and utilize current upto 1500 A.
On other hand, the nuclear reactor in power plant has performance of 950MW(950 000 000 W).
What I mentioned now were only maximal values, the energy required changes depending on how much load the device has to carry on.
The uncoupled locomotive will draw only 50kW from catenary, when attached to a long fully loaded freight train it will use its 4000kW like nothing.
The idle computer draws about 50W, when you turn on the games or heavy programs it may go as high as 1kW depends on the parts inside.
Coffee maker or light bulb are passive and they use same amount of energy everytime they are turned on.
If you need to know something about electricity feel free to ask but it's way too much to just write it here all
Or if someone knows better correct me if I'm mistaken.
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Most electrical power sources are run by systems that control the voltage. For example, in North America and many other places, the wall outlets provide 115 V as Alternating Current (AC). The utility company works hard to make sure that, no matter how much current is being pulled through their systems by thousands of users, the voltage in their wall outlets stays the same. When you plug in an appliance (say, a baseboard heater) its working component, the heating coil, has a fixed resistance. Thus, the current in Amps that can flow through that kettle's heater is determined by those other fixed factors: Amps = Voltage divided by current. Now, you will note that many baseboard heaters have visible heating coils that get red hot when operating. That is because the WORK being done by this electrical device is all being converted to HEAT and radiated out into the room. You MIGHT try to get even more heat (work) out of this heater by supplying it with an even higher voltage, and thus forcing more current through it. However, the coiled wire would get so hot the metal would melt and the wire would break, stopping the circuit completely. This is because the heater designers carefully made the unit so that, given just the right voltage supply, the heating coil would reach just the right temperature so that the rate of heat escaping the coil into the room exactly matches the rate of heat being generated in the coil by the electricity flow. So it is not a good idea to try to give that heater more voltage.Similarly, a motor might run faster if you fed it a higher voltage, BUT it also would generate a lot more heat inside its wire windings, and the heat would cause melting of metal and failure that way.
We measure electricity use in WATTS, and others have pointed out that Watts = Volts x Amps. Watts is a unit of the RATE of doing WORK, just as Horsepower is a Rate of doing Work. In fact, you can convert one to the other: 1 HP = 746 Watts. So OP's coffee maker (uses 120 V at 7 Amps = 840 Watts) uses a little over 1 Horsepower. Heating water takes work. To heat one gram of water by one Celsius degree takes one Calorie. It turns out that one Watt provides 0.2388 Calories per second. Since OP's coffee maker converts 840 Watts to heat, it can heat at the rate of 200 calories per second. Now, let's say the water put into the pot was at 20 C - room temperature - and it has to get to 100 C. So the heater has to raise the water temperature by 80 Celcius degrees. It can do that in one second for (200 / 80) = 2.5 grams of water. (In truth, the heater has to give the water lots more heat than that to actually make some of it boil, but that's more complication.) This is just to illustrate that the WATTS that came from electricity does some useful WORK on something - it heats the water.
Other types of electrically-powered machines do different types of work for us, but each always is just a machine to convert electrical energy into other forms of energy like heat and mechanical movement, at a RATE that depends on the Watts of energy being consumed. A higher wattage rating on a machine means it has been designed to consume and convert electrical energy at a faster rate; so, a larger motor may be able to power a larger fan and push more air around the room.
Now, on to car batteries. Ideally, a battery also is a source of a fixed voltage. In cars it is usually about 12.6 VDC. But no battery is ideal. In fact, each behaves like it has inside its case both an ideal battery that always provides a fixed voltage, PLUS (in series with it) a resistor that reduces the actual voltage available at the external terminals according to how much amperage is being pulled out at that moment. Then there's another complication. A battery stores electrical energy by using reversible chemical reactions. When it is charged after first being built, the charging current causes chemical reactions that use up a bit of its material to make new chemicals. These are stable and are the source of the voltage you can measure with a voltmeter. After that if you connect a load (say, a headlight) to the battery, the voltage will cause current to flow though the load. For that to happen, those same chemical reactions have to proceed in the OPPOSITE direction, regenerating the original chemicals in order to release the electrons needed for that current flow. When this happens, the supply of the "charged-up" chemicals is reduced, and hence the voltage is reduced slightly. If this continues for a long time, eventually the voltage will be reduced a lot, and we have to re-charge it to restore the battery to full voltage. That is why a car has a generator (powered by a belt from the engine) that provides a charging current to the battery as you drive.
OP, you asked why a car battery is rated in Amps? Good one, and the reason is that apparent internal resistance. Also, they are not being completely precise in making the statement. The specification is often for "Cranking Amps", and what it means is that during the use of the starter motor to start the engine, the battery will provide that many amps current to the starter while still maintaining a minimum voltage to the starter of (some particular number, probably about 8 to 9 Volts). You may have noticed that the Cranking Amps number is big - often 300 to 500 Amps - because that starter motor has to do a LOT or work to turn over the engine and get it going. If the battery is too small, or if it is so old it cannot perform as specified, when you try to use it the heavy current flowing through the starter motor will reduce the actual VOLTAGE getting to the starter and it won't get the required total power (Watts = Volts x Amps) to do that work. That is one of the consequences of that Ohm's Law thing that says Amps = Volts / Resistance. Re-stating that equation, Volts = Amps x Resistance. So, for that apparent internal resistor in the battery, the voltage drop across it is its resistance times the current being drawn. For a current draw of 300 amps, the battery terminal voltage could drop by 6 volts if the internal resistance is 6/300 = 0.02 Ohms. In such a case, in fact, the battery probably won't even deliver that 300 amps the starter wants - it will fall a some lesser point - and the starter won't turn over the engine fast enough to get it going. In fact, usually when this happens the reserve of "charged-up" chemicals in the battery is so low that its output decreases rather quickly, and the stater motor pretty soon stops turning at all. That's when we know we have a "dead battery" that won't start the car.
So, although a car battery is designed to deliver a fixed voltage to a load you connect to it, it has real limits on what it can do. And for us people who want the car to start reliably, an important other specification from the battery maker is how much real current it can provide under heavy load to turn over the starter motor (and hence the engine).
For people like me who live in colder areas (in the winter), you'll also see a specification for "Cold Cranking Amps". You see, the chemical reactions that go on inside a battery are affected by temperature - the lower the temperature, the less the output voltage - and hence, the lower the current it can push through the starter. So the battery makers also specify a Cranking Amps at a particular cold temperature so you can decide whether it is good enough to work in your cold winter weather.
We measure electricity use in WATTS, and others have pointed out that Watts = Volts x Amps. Watts is a unit of the RATE of doing WORK, just as Horsepower is a Rate of doing Work. In fact, you can convert one to the other: 1 HP = 746 Watts. So OP's coffee maker (uses 120 V at 7 Amps = 840 Watts) uses a little over 1 Horsepower. Heating water takes work. To heat one gram of water by one Celsius degree takes one Calorie. It turns out that one Watt provides 0.2388 Calories per second. Since OP's coffee maker converts 840 Watts to heat, it can heat at the rate of 200 calories per second. Now, let's say the water put into the pot was at 20 C - room temperature - and it has to get to 100 C. So the heater has to raise the water temperature by 80 Celcius degrees. It can do that in one second for (200 / 80) = 2.5 grams of water. (In truth, the heater has to give the water lots more heat than that to actually make some of it boil, but that's more complication.) This is just to illustrate that the WATTS that came from electricity does some useful WORK on something - it heats the water.
Other types of electrically-powered machines do different types of work for us, but each always is just a machine to convert electrical energy into other forms of energy like heat and mechanical movement, at a RATE that depends on the Watts of energy being consumed. A higher wattage rating on a machine means it has been designed to consume and convert electrical energy at a faster rate; so, a larger motor may be able to power a larger fan and push more air around the room.
Now, on to car batteries. Ideally, a battery also is a source of a fixed voltage. In cars it is usually about 12.6 VDC. But no battery is ideal. In fact, each behaves like it has inside its case both an ideal battery that always provides a fixed voltage, PLUS (in series with it) a resistor that reduces the actual voltage available at the external terminals according to how much amperage is being pulled out at that moment. Then there's another complication. A battery stores electrical energy by using reversible chemical reactions. When it is charged after first being built, the charging current causes chemical reactions that use up a bit of its material to make new chemicals. These are stable and are the source of the voltage you can measure with a voltmeter. After that if you connect a load (say, a headlight) to the battery, the voltage will cause current to flow though the load. For that to happen, those same chemical reactions have to proceed in the OPPOSITE direction, regenerating the original chemicals in order to release the electrons needed for that current flow. When this happens, the supply of the "charged-up" chemicals is reduced, and hence the voltage is reduced slightly. If this continues for a long time, eventually the voltage will be reduced a lot, and we have to re-charge it to restore the battery to full voltage. That is why a car has a generator (powered by a belt from the engine) that provides a charging current to the battery as you drive.
OP, you asked why a car battery is rated in Amps? Good one, and the reason is that apparent internal resistance. Also, they are not being completely precise in making the statement. The specification is often for "Cranking Amps", and what it means is that during the use of the starter motor to start the engine, the battery will provide that many amps current to the starter while still maintaining a minimum voltage to the starter of (some particular number, probably about 8 to 9 Volts). You may have noticed that the Cranking Amps number is big - often 300 to 500 Amps - because that starter motor has to do a LOT or work to turn over the engine and get it going. If the battery is too small, or if it is so old it cannot perform as specified, when you try to use it the heavy current flowing through the starter motor will reduce the actual VOLTAGE getting to the starter and it won't get the required total power (Watts = Volts x Amps) to do that work. That is one of the consequences of that Ohm's Law thing that says Amps = Volts / Resistance. Re-stating that equation, Volts = Amps x Resistance. So, for that apparent internal resistor in the battery, the voltage drop across it is its resistance times the current being drawn. For a current draw of 300 amps, the battery terminal voltage could drop by 6 volts if the internal resistance is 6/300 = 0.02 Ohms. In such a case, in fact, the battery probably won't even deliver that 300 amps the starter wants - it will fall a some lesser point - and the starter won't turn over the engine fast enough to get it going. In fact, usually when this happens the reserve of "charged-up" chemicals in the battery is so low that its output decreases rather quickly, and the stater motor pretty soon stops turning at all. That's when we know we have a "dead battery" that won't start the car.
So, although a car battery is designed to deliver a fixed voltage to a load you connect to it, it has real limits on what it can do. And for us people who want the car to start reliably, an important other specification from the battery maker is how much real current it can provide under heavy load to turn over the starter motor (and hence the engine).
For people like me who live in colder areas (in the winter), you'll also see a specification for "Cold Cranking Amps". You see, the chemical reactions that go on inside a battery are affected by temperature - the lower the temperature, the less the output voltage - and hence, the lower the current it can push through the starter. So the battery makers also specify a Cranking Amps at a particular cold temperature so you can decide whether it is good enough to work in your cold winter weather.
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