Help with Water to fuel to Water experiment
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Help with Water to fuel to Water experiment
Hi I was wondering how I could calculate the ideal voltage if I only used 3 batteries instead of 4.
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Re: Help with Water to fuel to Water experiment
Hello,
I apologize, but I am confused about the context of your experiment. Could you please provide more details, such as a link if it is based upon one of our own projects?
Sincerely,
Kayli Masuda
I apologize, but I am confused about the context of your experiment. Could you please provide more details, such as a link if it is based upon one of our own projects?
Sincerely,
Kayli Masuda
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- Posts: 2
- Joined: Sat Jan 27, 2018 4:13 pm
- Occupation: Student
Re: Help with Water to fuel to Water experiment
https://www.sciencebuddies.org/science- ... #procedure
I was wondering how i would calculate the ideal voltage as i only used 3 batteries
I was wondering how i would calculate the ideal voltage as i only used 3 batteries
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- Former Expert
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Re: Help with Water to fuel to Water experiment
Hi zapien2407,
Just out of curiosity, why 3-9V battery?
With 3 - 9V battery, you would be starting with 27Volts (3 * 9V). Assuming you are keeping the rest of the setup the same, I would suggest using the same ratio (30 /36) as the voltage being passed through the 10,000 ohm resistor. (resistors are linear devices and can be expected to behave the same in certain range.) So, in your setup, you will now have 27 * 30 /36 = 22.5V across the 10,000 ohm resistor. Following the technical note, the current is then 22.5V / 10000ohm = 0.00225 Amp or 2.25mA.)
From here, you should be able to follow the rest of the technical notes for your calculations.
Good Luck and please post again if there's anything else we can do to help!
Willey
Just out of curiosity, why 3-9V battery?
With 3 - 9V battery, you would be starting with 27Volts (3 * 9V). Assuming you are keeping the rest of the setup the same, I would suggest using the same ratio (30 /36) as the voltage being passed through the 10,000 ohm resistor. (resistors are linear devices and can be expected to behave the same in certain range.) So, in your setup, you will now have 27 * 30 /36 = 22.5V across the 10,000 ohm resistor. Following the technical note, the current is then 22.5V / 10000ohm = 0.00225 Amp or 2.25mA.)
From here, you should be able to follow the rest of the technical notes for your calculations.
Good Luck and please post again if there's anything else we can do to help!
Willey