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I am studying Electrical Engineering and I came across one definition of the volt being defined as follows:

V = J / C

Where "V" is voltage measured in "volts", "J" is energy measured in "joules" and "C" is charge measured in "coulombs". If this definition is correct, it would appear that a 10 V battery should be able to produce 10 joules of energy per coulomb of charge. My question is, what is the relationship between volts and Mechanical Energy?

The reason I bring this up, is because if we assume that the acceleration due to gravity is 10 m/s², the amount of energy required to move 1 kg of water a distance of 1 meter vertically is equivalent to 10 joules. I used the following formula:

F = m * a = (1 kg) * (10 m/s²) = 10 newtons.

W = F * d = (10 N) * (1 m) = 10 joules.

Does this mean that using a 10 V battery, I should be able to create a device that is able to move 1 kg of water a distance of 1 meter up in the air? I am trying to better understand the relationship of volts as a difference in Electrical Potential between two points in space and somehow using this potential to move objects with a certain mass a certain distance. Perhaps I am mixing different physics concepts together. Any insight into this would be appreciated. Cheers.

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  • \$\begingroup\$ Your first equation IS the answer to the question. No, a 10V battery cannot do the deed, unless a quantity of charge is removed from that battery. Solve for the C factor. \$\endgroup\$ – Whit3rd Jan 22 '18 at 21:28
  • \$\begingroup\$ You can also have mechanical potential. Water lifted 1 meter up in the air has mechanical potential. Since it has a mass of 1 kg, it has potential energy of about 10 J. If it had a mass of 5 kg, it would have a potential energy of about 50 J. Mass and charge are analogous. \$\endgroup\$ – davidmneedham Jan 22 '18 at 21:34
  • \$\begingroup\$ Perhaps look at the "electron-volt" unit to see where they are tied together. \$\endgroup\$ – pjc50 Jan 22 '18 at 22:15
  • \$\begingroup\$ You should be able to create a device that does that by using 1 coulomb of charge from the batery. You could also do it with a 5-volt battery using 2 coulombs, or a 20-volt battery using half a coulomb. Though FYI, the use of coulombs seems to be relatively uncommon, in this case we'd probably call them amp-seconds (and compute the time needed in terms of the current the battery is able to supply). \$\endgroup\$ – immibis Jan 22 '18 at 23:47
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Well yes, energy is energy. The only problem is converting it from one form of energy to another form of energy.

And you cannot take a random battery with 10 V. It has to have enough charge to be able to do the work.

So in a perfect world, you would build some magic mechanism, which would be able to lift something at 100 % efficiency when you supply it with 10 V. You want to lift 1 kg up 1 m, so you need approximately 10 J.

How much charge would you need? Well 1 C, but that's not a really useful measure you usually get with a battery. Most of the time you get something like amperehours or watthours. 1 Ah is equal to 3600 C (1 A is 1 C per second). So your battery would just need to have 1/3600 Ah. Or 2.8 mWh. This is so little energy - your phone could do this work surely 20000 times before running out of juice.

But your phone is running on 3.6 V - the voltage doesn't really matter (it will in a practical application), if you restate the equation, you will get J = V * C, so if you have less voltage you just need more charge to get the same energy.

Problem is, in reality you will have losses everywhere and you might get something like 80 % efficiency (conversion losses, friction in the mechanic) or something and so you will need more energy to do the work.

Devices like this do exist - they are called elevators.

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  • \$\begingroup\$ A water pump powered by three D-cell batteries can lift 1 kg of water to a height of 1 meter 3000 times according to its product description. \$\endgroup\$ – davidmneedham Jan 22 '18 at 21:51
  • \$\begingroup\$ @davidmneedham a typical D-cell has around 24 Wh, so you have 72 Wh to get 3000 J, which is rather horrible. I mean, that's like 1 % efficiency. I interpret the description a bit different though - it is able to pump 200 gallons per hour for five hours 40" up (I hate those units) So that's 1000 gallons 40" up. That would be 38500 J. Turns out to roughly 15 % efficiency - which is better, I don't have any figures what would be realistic for this. I would have guesses more in the 40's but well - that's guessing for you. \$\endgroup\$ – Arsenal Jan 22 '18 at 22:01
  • \$\begingroup\$ It's probably a good idea to point out the converse: A 1-V battery could also in principle lift 1 kg of water 1 m, if it had enough stored charge. \$\endgroup\$ – The Photon Jan 22 '18 at 22:09
  • \$\begingroup\$ @ThePhoton thanks, I used my phone example to include your point. \$\endgroup\$ – Arsenal Jan 22 '18 at 22:34
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a 10 V battery should be able to produce 10 joules of energy per coulomb of charge.

Yes. As you say, that's straight from the definition of a volt.

what is the relationship between volts and Mechanical Energy?

You already answered that yourself. The voltage is a measure of how much energy each Coulomb of charge has.

This has nothing to do with "Mechanical" energy. A joule is a joule whether that's a coulomb at 1 volt potential, A newton of weight raised 1 meter, or something else.

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Does this mean that using a 10 V battery, I should be able to create a device that is able to move 1 kg of water a distance of 1 meter up in the air?

If that battery can supply 1 Amp for 1 second ... yes. 1 Amp is 1 coulomb/second, and you dropped the coulomb from your actual question though it was, correctly, present in your introduction.

Connect it to a 100% efficient electric motor, winding string onto a drum, to raise the weight. 10V will translate to the motor running at a specific speed, and the torque required to raise the weight will draw a specific current (maybe 1 Amp) from the motor. If it draws less current ( 0.5A) that translates to less power (5W) , so your 10J work will take longer ( 2 seconds).

The trick, of course, is to find a 100% efficient motor, A small one like that (10W) is likely to be 50% at best, up to 90% above the horsepower range.

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