Gene0581 Posted March 10, 2019 Posted March 10, 2019 Hi- I am working with my son to create a homemade battery to power an old fisher price dune buggy that previously was powered by a 12v battery. We created a voltaic pile with pennies and zinc washers and it achieves 12 volts plus (we reduced the voltage to just under 12 volts). However, it doesn't power the motors on the buggy (or one motor). We were able to power the motors with a 9 volt household battery. What is the required amperage to power the motors (there are two- 15 teeth). Also, how do we achieve this with a homemade battery? Much thanks- need advice!
studiot Posted March 10, 2019 Posted March 10, 2019 18 minutes ago, Gene0581 said: Hi- I am working with my son to create a homemade battery to power an old fisher price dune buggy that previously was powered by a 12v battery. We created a voltaic pile with pennies and zinc washers and it achieves 12 volts plus (we reduced the voltage to just under 12 volts). However, it doesn't power the motors on the buggy (or one motor). We were able to power the motors with a 9 volt household battery. What is the required amperage to power the motors (there are two- 15 teeth). Also, how do we achieve this with a homemade battery? Much thanks- need advice! You need to look at the construction of your battery to reduce its internal resistance. Why is this in homework help?
John Cuthber Posted March 10, 2019 Posted March 10, 2019 You could try connecting many piles in parallel.
Sensei Posted March 10, 2019 Posted March 10, 2019 (edited) Like John mentioned, with limited explanation, we have two things to bother, voltage at required minimum, and current at required minimum. P = U * I If you have required voltage U (e.g. 12 V), but current I is too low, you have too low power.. Batteries connected together in serial have increased voltage. Batteries connected together parallel have increased current. Batteries connected together both serial and parallel have increased either voltage and current, thus the higher power. Edited March 10, 2019 by Sensei
studiot Posted March 10, 2019 Posted March 10, 2019 (edited) 2 hours ago, Sensei said: Like John mentioned, with limited explanation, we have two things to bother, voltage at required minimum, and current at required minimum. P = U * I If you have required voltage U (e.g. 12 V), but current I is too low, you have too low power.. Batteries connected together in serial have increased voltage. Batteries connected together parallel have increased current. Batteries connected together both serial and parallel have increased either voltage and current, thus the higher power. Perhaps the homework lesson here is that the power requirements cannot be met by home constructible primary cells. That is why, even, commercially, storage batteries are used, preferably rechargeable ones. One alternative might be one (or more) of those new fangled capacitors in the farad range. Edited March 10, 2019 by studiot
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