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Data: PD=12.0 V

q = 100,000 C

W= ? (need answer in Joule)

Basic Equation: Pd= W/q voltage(PD)=Work(W)/charge(q)

Working Equation: W=PD(q)

Substitution: W= 12.0 V(100,000 C)

W= 1,200,000 J

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12y ago
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13y ago

No. The voltage from a 12 volt battery will vary from over 14 volts down to near zero (at discharge). An ampere delivered at X volts will have X joules. A fully charged battery which is still on charge (alternator running or 117 volt powered charger connected, for example) could be delivering 15 joules per ampere, or even a little more. An older battery, starting a cold engine, could drop to 6 volts, thus 6 joules per amp. Batteries can go down to near zero under load, but below about 6 volts, the engine is not likely to start. Keep in mind that the joule is a coulomb-volt, a unit of energy. So we are getting a coulomb per volt-amp every second. This is an energy (a rate) of 1 watt per volt-amp. So 100 amps at 10 volts would be 1000 watts. This is doable by a good, charged, lead-acid battery. This would be 1000 joules every second. On a really cold day, we might get 250 amps at 8 volts for 2000 watts. This is 2.68 horsepower and should start our car.

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10y ago

Welll ... they don't actually flow 'through' the battery. It's much more

common to talk about the coulombs that flow through the external circuit,

and the energy they give up along the trip.

Each coulomb of charge that flows out of one terminal of a 12v battery,

then through an external circuit and back into the other terminal, yields

12 joules of energy.

1 volt = 1 joule per coulomb

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Q: How much energy is supplied to each coulomb of charge the flows through a 12-v battery?
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