Unmatched voltages on a transformer's primary coils

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In summary, a transformer will convert the power from one voltage to another. The two coils on the transformer will produce different voltages and currents. If the two coils are of the same size and number of windings, then the two generators will add together and create a higher output voltage. If the two generators are of different sizes and number of windings, then the two generators will subtract from each other and the output voltage will be lower.
  • #1
infamous_Q
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Ok. say you have a transformer hooked up to two AC generators. They each produce the same amount of voltages. For the primary coils on the transformer (the 2 generators..) one is coiled 100 turns and the other 10 turns...the result?

Continuation of question...
now assuming that this means one source will be transformed to have a high voltage and low current. and the other will have a high current and low voltage, this would mean that they would "fight" each other. But what would this "fight" result in? Loss of power in the form of electricity? or mechanical resistance (from teh generator spinning)? I have no idea to be honest...so any help for these questions would be greatly appreciated!
 
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  • #2
infamous, we talked about this in PMs and I'll try to use a mechanical analogy again to make this easier.

Instead of a transformation of electrical-magnetic-electrical as a transformer would perform, think electrical-airflow-electrical where we have an electic motor fan that is blowing against a wind generator.

Our input is a certain amount of power that can be measured as a voltage (pressure) and current (amount flowing). When its converted to wind we have the same qualities of a pressure and amount flowing that equate to a certain amount of power. We could measure the pressure and amount and know how it will influence the wind generator.

Now you are asking about using two different sized fans. If all things are kept equal like wire size and so on either the small fan is going to burn up and/or the large fan will spin very slowly. Each fan should be optimized to make the best use of its available power.

If we have X volts to work with, then the large fan we apply a large current we'll call Y, giving us an input power of XY. Using X volts on the small fan we'd normally need less power and a smaller current we'll call Z, giving us an input power of XZ. So we know that XY > XZ, and our input power is XY + XZ.

Now you decide to change the smaller fan to be more powerful by angling of the blades or making it spin faster and now it handles XY power too. Our input power is now 2XY with both fans running. But now the small fan needs to move an aweful lot of air to match the power of the large fan and efficiency likely roars its ugly head. This is likely not as efficient as just using two of the same optimized fans to begin with since it is designed to work with the appropriate level of power and you're not fighting ending up on a poor part of an efficiency curve like trying to spin a small fan super fast.

Because with a transformer the number of windings and size of wire is very important to ensure the power is delivered properly and not wasted as heat. So its not so much a battle amoungst the two coils but within each coil itself and its ability to transfer the power efficiently. And that is where the choice of wire size and number of windings is figured out and should match the application.

And even if the phase isn't maintained (fans aren't blowing in the same direction) there is no mechanical fighting per se, its more of a cancellation determined by how much of an angle difference there is in the directions.
 
  • #3
so if using two generators...and one small winding and one large winding. rather than have the one generator try to spin slower per se...the electricity will cancel itself out? so since power in two primary coils add together (50W + 50W will produce 100W in the secondary coil). would they subtract each other? or turn one sort of "negative"?
 
  • #4
Spinning the generator slower creates all sorts of problems. Change only one thing at a time!

If both sources are in-phase operating at the same frequency then they will simply add like constructive interference. (see link below)

However, assuming they are identical generators (and everything else is the same except windings), using different windings will cause massive problems for the lower turn one - the winding needs to have a certain number of turns in order to not work as something close to a dead short. Nothing wants to encounter a short! If the lower number was sufficient and then the larger turn number was well oversized its efficiency would likely be lower - now entering the art of transformer design and some handy tables would be nice to have access to...

If the two generators were operating at the same frequency but were 180 degrees out-of-phase and powering identical windings then the magnetic fields would cancel each other out since it would be destructive interference.

http://www.glenbrook.k12.il.us/gbssci/phys/Class/waves/u10l3c.html
 
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  • #5
ok...so like in the waves it's constructive interference, just the size of teh waves are different this time around. what happens then? like what problems actually occur?
oh, also, since this is interference, is it the overall power that interferes or the voltage and current interfering with each other seperately?
 
  • #6
Simple physical analogy to explain this:

You push a little on a chair next to you. You push but it doesn't move so you perform no work on the chair. This is like having a voltage (pressure or push) but no current flowing, and with zero current you have zero power.

You push the chair and it moves along at 1 cm/s because the force you apply is equal to the force of friction from pushing the chair. If you push harder you could push a second chair or push the original faster and so on. But the idea is that your pushing force and the amount of friction dictate the speed the chair moves along and thus the rate at which you're doing work - power.

In electricity its all dictated by ohm's law: Voltage = Current * Resistance

Voltage = E
Current = I
Resistance = R

E = I * R

(c is used for the speed of light, v is used for velocity, E&I were used instead)

Power = Voltage * Current

And simple algerbra let's you see how you can solve and see that for example power=resistance * current^2 and how that let's you know only the resistance and current and still know the power (you could also find the voltage). But regardless, the voltage, current, and resistance are all tied together.
 
  • #7
so then...since Pt would technically equal (in its most basic form..) P1 + P2. then Pt would also be (V1+V2)*(I1+I2)?
 
  • #8
No, you have a typo.

P1 = V1 * I1
P2 = V2 * I2

Pt = P1 + P2
Pt = (V1 * I1) + (V2 * I2)
 
  • #9
aaah thanks.

Edit:

would the electricity in the wire look like V1+V2 for teh voltage, and I1+I2 for the current?
 
  • #10
Well, now you need to know how things are hooked up. If this were in series with a simple load it could really be that easy. Like if you measured some AA flashlight batteries it would work this way.

But if you have just one wire and two sources you would have to figure it out using something like kirchhoffs laws of voltage and current in a circuit. And that's just linear relationships, something like a light bulb has a resistance that changes as it heats up so as you flow more current this can change a bunch.
 
  • #11
ok...so say that these two coils on teh generator power the single secondary coil, which is hooked up to a parallel circuit...
 
  • #12
Ok, what happens at the secondary of a transformer is heavily influenced by what's happening at the primary but they are isolated and separate systems. That's why I used an electric fan and wind generator as metaphors to describe how the two are not directly coupled but instead coupled by air movement. Inside a transformer its a changing magnetic field.

So the math could be simplified to Power in and Power out to find the maximum voltage/current available to your load if you had an ideal transformer (and maybe oversized to allow for regulation and so on).

Inside a transformer the Primary's voltage is converted to electromotive force (Gilberts or ampere-turn) and the current is converted to flux (Maxwell or Gauss) and the secondary converts those back to voltage and current. Its like a torque converter in an automatic car transmission, it converts the torque and RPM to a higher torque at a lower RPM to make it easy to get the car moving (along with slipping to make for smooth starts).

Then the secondary could be thought of as a voltage source to the load with a maximum current capacity of the ideal transformer. So you know its voltage and with the resistance (technically impedance) you can find its current.
 
  • #13
Ok..that makes way more sense, because the changing magnetic field in the transformer induces a current in teh other coil. So what makes a highvoltage magnetic flux different from a high current flux?

and what would happen if (in parallel or series even..) you had an external power source (like a battery or something) where:
for parallel: the voltage was higher than teh transformers output
for series: the current was higher than the transformers output
what would happen in the transformer?
 
  • #14
high voltage = high EMF 'push'
high current = high flux 'amount'

(assuming you don't saturate the core, and if they are both high you're simply pushing around a lot of magnetics)

Ok, so you're going to hook a battery up to the secondary - Let's assume you have a rectifier bridge to convert AC back to DC then.

In parallel the higher voltage feeds into the other. This would be typical, its how you charge a battery by having just a slightly higher voltage and why the battery will get warm as it sinks that power. Which means the load will likely not see too much higher of a voltage then somewhere between the output and the battery's resting voltage.

In series its no simpler - the lower current wins since the current is going to be the same throughout the circuit so the bottleneck is the predominant (probably ESR limited). Of course, in series the voltage is higher so with the same load the current is going to try to be higher and the power has gone up by the product (like a square power) and the fuse blows and/or the components let out their magic smoke. Likely 4:55PM on Friday when the boss has stopped by to chat and you're hungry to boot. :smile:
 
  • #15
ok...so if you had one primary coil with a high voltage and one with a high current youd just have a lot of nothing?

as for the second thing what if you hooked it up to a capacitor? (one with a high voltage and relatively high current)...i don't completely understand capacitors though..and not at once either. Consider 2 generators...in different phases. (ie. since they basically produce power in a save wave form...we'll use 360º as a full revolution), one produces a high voltage from 0-180º and is disconnected until after the full revolution...the other one produces a high current from 180-360º and does nothing before that...would it charge the capacitor with a high current and voltage so taht the load can then take the power from the capacitor?
 
  • #16
Whole lot of theory and not much practical here...but here goes nothing.

On the first one, it depends. Its all about the power. If you have 200V and 1mA for one and .1mV and 200A for the other you could light up a few LEDs or a tiny auto light bulb with that tiny amount of power.

Remeber that voltage and current are connected, you need one to have the other. If you have the same load, if you have high voltage you have high current. If you have low voltage you have low current. You don't get one or the other, you always get both. You could change the load to suit the proportions (or a transformer to make them work out better) but there is no free lunch. If you want high power and high current you need lots of power, about the only way around it is to take a smaller power applied over time and store it like in a battery.

A capacitor only works on DC for storing power that could be applied to a load. And even a capacitor the size of a soda can could only keep a small flashlight bulb going for a minute or less, so its far less space efficient than a battery by maybe 100x or more.
 
  • #17
but its not liek i could build a battery that keeps recharging through a dc input WHILE mainting power to the load (from teh battery)...could I?

and yes I am aware it's lots of theory..which is why i came here to ask for help, since i (obviously) lack much knowledge in these departments.

Also, would this battery (the one that could be rechargable..if that even works out) be able to hold up the charge? (high voltage low current in, then high current low voltage in, to build up a source with high current and high voltage?)
 
  • #18
infamous_Q said:
but its not liek i could build a battery that keeps recharging through a dc input WHILE mainting power to the load (from teh battery)...could I?

No, but a capacitor works in a similar manner and has its own challenges with charging/discharging (namely the amount of current available).

infamous_Q said:
Also, would this battery (the one that could be rechargable..if that even works out) be able to hold up the charge? (high voltage low current in, then high current low voltage in, to build up a source with high current and high voltage?)

Ok, new analogy.

You have a baseball glove on and someone throws you a baseball and as you catch it the kinetic energy is absorbed into your body. Someone could fire a BB gun and deliver the same kinetic energy, someone else could lighlty throw a 8lb steel shotput ball into the glove and again deliver the same kinetic energy. Can your glove handle all three? Or would you rather have a design optimized for each situation (I'll assume you agree).

I dislike water analogies because they don't obey closed circuits and so on but it'll work well enough here.

Your capacitor is a very tall water tank like is used for city water distribution. 1st scenario you have a pump that can pressurize a garden hose to a high enough pressure to fill the thing - it can actually fill it but it'll take a while. 2nd scenario you have a swimming pool - it has no pressure but could fill up the first 6ft very quickly if the elevation allowed it. So it does very little good to fill 6ft of a tall tank. And it'll take forever to fill it with a garden hose. Instead you need a pump and engine (power) like a firetruck to fill it.

And that is how a city water system works - they pump water into the tower all day long 24/7 but they have planned for people to use water in certain amounts at certain times of the day that average out to a value close to the amount they pump in on average. The tower allows for short bursts of higher flow than the pumps may be able to deliver, but this must be followed by a period of low usage to allow for a recharge. When averaged out over a day, its the same amount.

The big problem is that capacitors store electricity in a very inefficient manner. Its like a fire truck - it may have 1500 gallons of water on board and that may be all the truck can handle for weight, but it can pump all that water out in just over a minute. It takes a lot longer than that to put out a burning house. So they need the fire hydrant nearby to give them a supply of water (or use it in a more efficient manner) because there is no escaping the need for sheer volume.

And if you need continuous large amounts of power you need a source that can provide it. Period.
 
  • #19
and what if the capacitor is constantly being charged and discharged. BUT the discharge is lower than the max the capacitor can hold (most of the time...it varies) and the charge going in is constant?
 
  • #20
It'd be no different than if the capacitor wasn't there except the ripple (amount voltage varies) would be larger without the capacitor. That's why cities use water towers, so you get constant pressure (and so you have water even if the power goes out). To get current out of a capacitor the voltage MUST drop and it drops like a rock - that's why you need a capacitor 100x larger than a battery to light up a small light bulb, it runs out of voltage very quickly as it discharges. A whole bunch of charge in a capacitor doesn't do anyone any good if it doesn't have any voltage to push it out.

And the capacitor only works on DC like a battery. Electricity generation and transport is almost all AC so its easy to step up with a transformer to avoid heat losses from sending current and then stepped down to use. Its also easy to convert AC to DC and without any complex components. Edison lost the battle for DC a very long time ago and with good reason. Now capacitors are used in power transmission to correct the phase angle (current lags voltage) of the AC power but the capacitor doesn't really store anything.
 
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  • #21
ok...so:
if used in DC the capacitor CAN act like a battery. BUT no matter how many amps there are stored up (i think i got this right...) without even some sort of voltage, the capacitor won't power anything. Correct?
So then if you had (back to this multiphase idea) 1 transformer (primary coils n what not don't matter for right now...or the # of generators) pumping out something like X amps...but a low voltage, the capacitor would at least become charged with X amps but it wouldn't make it to the load. Then in the next phase with a high voltage and low current (we'll call it Xb) going to teh capacitor, the capacitor would charge up to X+Xb amps, and would have a relatively high voltage so that it could reach the load, right? (i know there's a flaw in there somewhere..especially because even though X amps is present there's still a low voltage to push it..)
2) what if you hooked up a high current source and a high voltage source (different sources) to a capacitor?
 
  • #22
Ok, a capacitor could be thought of as like a battery that goes dead super super fast, typically its used because it has a low ESR (equivalent series resistance) and doesn't require maintenance.

And situation 2 is a good way to blow something up and does nothing otherwise.

A capacitor works because you are stuffing an abundance of electrons on a plate of material. You can have all the current you want, without voltage it doesn't charge the capacitor. If your house was on fire would you want a garden hose hooked to the city water system or a lake nearby? At least the hose has pressure to get some water on the fire, that lake doesn't do you any good just sitting there.

You need both voltage and current. That is why we use the term power. A firehose has a lot of flow and a lot of pressure and that's how it gets the job done.

A capacitor stores a charge and reflects this. Its an charging curve based on the natural log e, here's a quick example with a 100V cap.

1 sec, 64V, 40% power
2 sec, 86V, 75% power
3 sec, 95V, 90% power
4 sec, 98V, 97% power
5 sec, 100V, 100% power

Notice how if you only drop 2V then you only get 3% of the power, if you drop 5V then you only get 10% of the power. If you have a light bulb its going to be dimming at the 5V drop point. And try to find a commercially available capacitor big enough to run light bulb like the one lighting the room you're in - it would be bigger than your desk!

A 60W headlight produces light similar to a 60W light bulb in a house even though one uses lower voltage/higher current and the other uses higher voltage/lower current. You're not going to gain anything with high X or high Y or any combination of that. You need power, I'm not sure how to explain it any simpler than that. :frown:
 
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  • #23
lol i get what your'e saying..im just trying to find away around it...i know its very unlikely but hey so was going into space at one time. So...i have a few more questions:

if you have say 64V going into a 100V capacitor will it charge up to 100% ever? (the others will follow later)
 
  • #24
Going into space simply required finding the right optimization of many many parameters. Disobeying the laws of physics is different.

That's what the capacitor would be at if fed 100V continuous through a resistor and you disconnected it to measure it. When you're charging a cap its nearly a dead short from a DC perspective. If you had enough power and a cap with a low enough ESR it could charge in .001 of a second. Except that would require a steady voltage and massive current and that level of current would likely heat it enough to make it boil and/or explode.
 
  • #25
hmmm...my ideas are quickly running out here...ill post again l8er
 
  • #26
ok. i have a few more questions...
1) capacitance = (current x time)/voltage ...correct?
2) what if you had a high current source quickly charge up a capacitor (no load on the circuit, just the capacitor, and only charged to like 90% or so), then hooked it up to (while disconnecting it from the high current source) high voltage/low current source (so not to over charge it) to finish charging it, or at least to about 98% charge. What would happen if you then cut it off from the power source charging it, and plugged it into a circuit? OR what if you just hooked it up to the high voltage source and a circuit w/ a load at the same time?
 
  • #27
1) Yes
2) Why would you go through all that trouble? Simply connect a capacitor up to a voltage source and the cap will charge. No need to switch from a current source to another type of source. As long as the voltage source doesn't exceed the voltage rating of the cap, the cap will be fine---though it is a good idea to use higher voltage ratings than those present in your intended circuit. As for what would happen: the cap would discharge--it would act like a battery. This fact is used in many applications. Automotive computers use caps as battery backup sources because real batteries have a shorter life span than a capacitor. I've seen field test equipment run for weeks off of a charged cap.

Think of a capacitor as a battery. What happens if you place a battery in series or parallel with another battery of greater or lesser charge (four scenarios for you to wrap your mind around). The same thing happens to a cap. If the source voltage is greater than the cap voltage, the cap will charge. When the cap is fully charged, the current through the cap drops to zero and the load will be supplied. If you disconnect the source the cap will discharge through the load. Depending on the size of the cap and the load's current requirements, the cap may discharge almost instantly or it may take weeks (most car computers will keep the volatile memory intact for 20 minutes to an hour after the loss of battery power).

Here, a basic law in physics(and EE) is that you can't get something for nothing. A fundamental observation from thermodynamics(and physics and ME and EE yadda, yadda, yadda) is that all processes incure losses. Can you charge a cap to 100V from a 64V source? Yes, voltage multipliers do exist; however, the total power out will be slightly less for the circuit using the voltage multiplier because the support cktry needed to boost the voltage consumes power. If you don't need a higher voltage then don't step it up because doing so incures losses---your system becomes less efficient. If you want to make 120V it's better to do it at the generator then afterwards using add on circuitry.

As for your capacitor inquaries: caps are batteries. They charge, they discharge, they store energy---they just happen to be frequency dependent as well(very useful for things like filters). Think in those terms and the answers to your questions should be apparent. In fact, if you look at the graph for a battery charge it should look exceedingly similar to that af a charging cap (ignoring the portions of the charge graphs where variables are controlled i.e. maintaining constant voltage by adjusting current).

So, what would happen if you charged a battery and connected it to a circuit?
 
  • #28
the battery would output power and the load would be supplied, however it would probably last longer than the battery (but that depends on a few things).

as for my scenario i asked this because with a high current you could charge teh capacitor much more quickly than with a low current (i think). Then with the high voltage connected to charge it a bit more, and then you could hook it up to the high voltage source to charge the voltage to full capacity (i think I am getting this right...but there are more questions ahead) and then you could hook up the capacitor to the load so the load could be supplied with as much the voltage and current needed relatively quickly.

Here's the other question to make sure I am understanding stuff correctly (or rather set of questions):

Say we have a capacitor of 1 farad (yeah i know they're big..this is just an example), which is rated at 30V. Now since C = (A*s)/V, 1 = (A*s)/30. SO A*s must equal 30. Which leaves a lot of room for different variables. Again, if I am getting this right this means that the capacitor could come to full charge if the current was 30 amps after 1 second, OR if the current was 15 amps after 2 seconds. This would mean it could discharge at teh same rate correct? BUT what if it was charged at under 30V at a current of something more like 300A (i know its not likely) for .1s?
 
  • #29
infamous_Q said:
the battery would output power and the load would be supplied, however it would probably last longer than the battery (but that depends on a few things).

as for my scenario i asked this because with a high current you could charge teh capacitor much more quickly than with a low current (i think). Then with the high voltage connected to charge it a bit more, and then you could hook it up to the high voltage source to charge the voltage to full capacity (i think I am getting this right...but there are more questions ahead) and then you could hook up the capacitor to the load so the load could be supplied with as much the voltage and current needed relatively quickly.

Here's the other question to make sure I am understanding stuff correctly (or rather set of questions):

Say we have a capacitor of 1 farad (yeah i know they're big..this is just an example), which is rated at 30V. Now since C = (A*s)/V, 1 = (A*s)/30. SO A*s must equal 30. Which leaves a lot of room for different variables. Again, if I am getting this right this means that the capacitor could come to full charge if the current was 30 amps after 1 second, OR if the current was 15 amps after 2 seconds. This would mean it could discharge at teh same rate correct? BUT what if it was charged at under 30V at a current of something more like 300A (i know its not likely) for .1s?

No, the battery will not output power if the source is at a higher potential. The battery becomes a load. If the battery were able to output power against a source of a higher potential then we would not be able to charge batteries at all.

Current has nothing to do with the rate of charge(to an extent) T=R*C (from ohm's law we know that E=IR---current in a circuit depends on the resistance of the circuit). While 1) above is true, it hides the simple fact that current depends on voltage (the forward push) and circuit resistance(the push in the opposite direction).

I hate to say this, but you really need to investigate the inner workings of capacitors (and inductors). You don't need to look into the diff eq aspect because RC circuits are fairly easy exponentials(ignoring the inductive portion : The generator). Moreover, it seems you may need to take a couple of steps back and (re)learn basic circuit theory i.e. what happens when you have mismatched voltage sources.

Now, I'm going to answer your last question with a question? What would happen? What would the graph look like? How many time constants does it take to charge a capacitor? What does the I(t) through a capacitor look like? You can come up with as many scenarios as you like; however the operation of the capacitor will not change. You won't get more power out(you WILL always get less). You can utilize a cap to produce huge current spikes(inductors will do this as well) however, the duration of the spike is inversly proportional the the magnitude of the spike.
 
  • #30
ok. so then here's a simpler question: how long does it take a capacitor to charge on a circuit with no load?
 
  • #31
infamous_Q said:
ok. so then here's a simpler question: how long does it take a capacitor to charge on a circuit with no load?

Please rephrase your question. Do you mean connect a cap to a source only?

Since [tex]I=\frac{V_b}{R}e^{-t/{RC}}[/tex]

your question is meaningless.

the actual answer to your question is somewhere between 4.6 and 5 [itex]\tau[/itex] depending on how accurate you want to get (you can go greater than 5[itex]\tau[/itex] if you like but you won't see any difference really).

[edit] the above is for a series RC circuit. A parallel RC circuit would be along the same lines though. The charge on a cap follows an exponential function and is considered fully charged at 5 time constants.
 
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  • #32
infamous_Q said:
as for my scenario i asked this because with a high current you could charge teh capacitor much more quickly than with a low current (i think).

No, you can have all the current you want. Once the capactior is at the supply voltage no more current will flow. So the capacitor charges by VOLTAGE and the current is determined by the resistance (see ohm's law).

You are continously posting about getting high voltage and current and are trying to cheat the power out of some switch scenario, but that doesn't work!

If you charge up a cap it takes time yes, and you could figure how many joules that is (or just use 1/2 C * V^2) and that is what you'd have available at discharge.

Yes the shorter time of discharge the larger current you can extract, but as you extract the current the voltage will drop like a rock (just like on the charge curve the discharge follows a similar path). And even if you get 8000A out of some big cap, it doesn't do you much good if it lasts 1msec does it? Unless you're doing an experiment in lightning or a giant camera flash or something like that, most loads like a power source. And regardless of how you skew the equation, you can only expect to extract the same amount of joules you put in.
 
  • #33
ok...so since I am wrong on everything thus far and my idea(s) won't work...i have a few final questions (for now at least..): say you have an AC generatore, with a relatively low resistance in the circuit (either in the form of an AC circuit, or a DC circuit which uses a converted AC generator for the power source) and you have a voltage of 50 V and a Resistance of 2 ohms (i know its not going to happen..just a question as usual)
1) does Ohm's law apply here? as in...will the circuit actually have 25A runinng through it?
2) and what if the generator can't supply that much mechanical power to be converted? (i'm going to assume that the voltage would simply drop until both values were optimal, i'd just like to know whether I am right or wrong on that)
3) (if i guessed right on #2) what if the power is transformed before hand, will anything change?

and btw, thanks for coping with my idiocy everyone, I am still kinda waiting to graduate high school so i can go get an electrical engineering degree.
 
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  • #34
1) If the generator can supply that much power and if we're talking about shorting the generator to ground then the answer to your question is yes.

2) Yes, the generator would slow down to the point where the mechanical input equals the electrical output---Essentially, Power in will equal power out.

3) No. Nothing will change. Power in will still equal power out. You can't cheat death and you can't make power from thin air. In fact, if you do a conversion then the electrical power out will be less than the mechanical power in because of added losses in the conversion cktry. Conversions=loss of usable power.
 
  • #35
ok thanks...guess what though, i have another question for you helpful people.

is it correct that emf produced in a coil is proportional to (or the equation is...):
emf = 2pi * number of coils * magnetic field coil is in * frequency coil's turning at * surface area of the coil
 

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