Volts/ Amps / Watts calculation
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A little while since I did my Physics O level........
A transformer steps up 2*AA Batteries (2*1.5v) to 2,700 volts, no idea on the amps, any idea of the wattage at 2,700 volts ?
D
A transformer steps up 2*AA Batteries (2*1.5v) to 2,700 volts, no idea on the amps, any idea of the wattage at 2,700 volts ?
D
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You need amperage
power = current * voltage
For a transformer that is X% efficient
Current in * voltage in = (X/100)(current out * voltage out)
power = current * voltage
For a transformer that is X% efficient
Current in * voltage in = (X/100)(current out * voltage out)
Last edited by Rage!; 13 October 2006 at 08:04 AM.
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P=IV or I(sq)R . I=V/R (I think) so if you have the resistance of the transformer then you should be able to get the power as well?
No, thinking about it you need to have an output current first. Or a resistance of what you're outputting to. Maybe.
Tell us a bit more? PSL claims to be a nuclear scientist - he should be able to answer that straight away...
SB
No, thinking about it you need to have an output current first. Or a resistance of what you're outputting to. Maybe.
Tell us a bit more? PSL claims to be a nuclear scientist - he should be able to answer that straight away...
SB
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Originally Posted by Dunk
Thks - so what would a 2* normal (everyready / duracell etc.) AA battery put out ?
D
D
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A transformer would only step up an AC current; batteries are DC.
The primary winding of the transformer would look pretty much like a short circuit to the batteries, so they would discharge themselves smartly.
You would need an invertor or similar DC to DC convertor in order to step up the output from a battery. This would then have its own power handling capacity which would dictate the current available and hence the wattage.
The current available from a battery depends mainly on its internal resistance (sorry, don't know the figures off hand for a 'standard' AA battery), but isn't it pretty irellevant due to my first statement?
The primary winding of the transformer would look pretty much like a short circuit to the batteries, so they would discharge themselves smartly.
You would need an invertor or similar DC to DC convertor in order to step up the output from a battery. This would then have its own power handling capacity which would dictate the current available and hence the wattage.
The current available from a battery depends mainly on its internal resistance (sorry, don't know the figures off hand for a 'standard' AA battery), but isn't it pretty irellevant due to my first statement?
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And using the old adage, "Power in = Power out", which is true for a transformer, (forgetting wasted power in the transformer, for the moment), the output WATTAGE will be the SAME as the input WATTAGE.
ie: as output voltage rises, output current falls, such that voltage x current will be a constant, (given by input voltage x input current.)
All this assumes the source CAN be transformed, ie, is AC![Wink](https://www.scoobynet.com/images/smilies/wink.gif)
Alcazar
Actually, you CAN get a transformer to step up, or down DC, now I think on, BUT, you have to work for it:
You can do it by "flashing" the connections to the battery across the input. You'd get a similar flash of output, at a higher, (or lower), voltage.
Repeatedly "flashing the connection" will produce repeated flashes of output.
You can use a make and break circuit, like that on an electric bell, or buzzer, to repeatedly switch on, then off, the input. You get repeated bursts of output, timed with the make and break.
You could use a mechanical device to open and close contacts on the input. This is similar to the old contact-breaker ignition.
HTH Alcazar
ie: as output voltage rises, output current falls, such that voltage x current will be a constant, (given by input voltage x input current.)
All this assumes the source CAN be transformed, ie, is AC
![Wink](https://www.scoobynet.com/images/smilies/wink.gif)
Alcazar
Actually, you CAN get a transformer to step up, or down DC, now I think on, BUT, you have to work for it:
You can do it by "flashing" the connections to the battery across the input. You'd get a similar flash of output, at a higher, (or lower), voltage.
Repeatedly "flashing the connection" will produce repeated flashes of output.
You can use a make and break circuit, like that on an electric bell, or buzzer, to repeatedly switch on, then off, the input. You get repeated bursts of output, timed with the make and break.
You could use a mechanical device to open and close contacts on the input. This is similar to the old contact-breaker ignition.
HTH Alcazar
Last edited by alcazar; 13 October 2006 at 04:22 PM.
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Originally Posted by Sbradley
.......... PSL claims to be a nuclear scientist - he should be able to answer that straight away...
SB
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This has some internal resistances for AA batts:
![](http://home.att.net/~mikemelni1/intresa.GIF)
So assuming 0.5 Ohms, an AA should be able to produce about 3 Amps maximum (if my maths is correct) at 1.5 Volts.
Two of them in series would still give this current I think, but at 3 Volts.
This would be 4.5 Watts.
Assuming DC to DC conversion at 100% efficiency (which is, of course, impossible), this would result in 4.5 Watts at 2,700 Volts, i.e. 1.667mA.
So assuming 0.5 Ohms, an AA should be able to produce about 3 Amps maximum (if my maths is correct) at 1.5 Volts.
Two of them in series would still give this current I think, but at 3 Volts.
This would be 4.5 Watts.
Assuming DC to DC conversion at 100% efficiency (which is, of course, impossible), this would result in 4.5 Watts at 2,700 Volts, i.e. 1.667mA.
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As others have said you can't easily get a standard transformer to step up DC.
The load on the primary will vary depending on what load is connected to the secondary of the transformer and what losses the transformer has at that given load.
Just be carefull what you are doing, electricity can be fatal at even very low currents, even currents as low as 30milliAmp can be fatal.
I've just worked nearly 60hours in the last three days, a small fault on an 11000V switchboard ionised the air and broke down the insulating airgap between the three phase busbars. The symetrical short circuit and 126MVA (MegaWatt ish) fault current was enough to cause a massive explosion which took out half the 30metre board.
Cheers
Lee
The load on the primary will vary depending on what load is connected to the secondary of the transformer and what losses the transformer has at that given load.
Just be carefull what you are doing, electricity can be fatal at even very low currents, even currents as low as 30milliAmp can be fatal.
I've just worked nearly 60hours in the last three days, a small fault on an 11000V switchboard ionised the air and broke down the insulating airgap between the three phase busbars. The symetrical short circuit and 126MVA (MegaWatt ish) fault current was enough to cause a massive explosion which took out half the 30metre board.
Cheers
Lee
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With my electrical knowledge I couldn't tell you if it's a transformer, transducer or transvestite, but in an electric fly swatter it takes 2 AA batts & steps them up to 2,700V.
Many thks for the input, sounds like <5 watts is the answer.
D
Many thks for the input, sounds like <5 watts is the answer.
D
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