D)
A
battery of potential 12V (voltage) gives each charge of 1C a potential
energy of 12J. WHile flowing in the circuit, the charges lose
their energy because some device will transform the potential
energy into another form of energy. All these devices have a resistance
noted R. THe energy lost per second
(Electric power) is proportional to R. In the same way, the water has a
potential energy because it is raised above ground level, the initial
energy is spent has it go through the pipes and encounter resistance.
The
battery gives each charge of 1C potential energy. The potential energy is
converted to another form of energy. The conversion is never 100%
efficient. Some energy will converted to thermal energy. (A bulb gets
hot, the wires get hot, the battery gets hot ...). Once the charge have
to more energy, they flow through the battery (the pump here) and get a
new boost.


| Current (A) | Potential Difference (V) |
| 0.01 | 2.3 |
| 0.02 | 5.2 |
| 0.03 | 7.4 |
| 0.04 | 9.9 |
| 0.05 | 12.7 |

source: http://www.pschweigerphysics.com/dccircuits.html
Here is an example. Suppose we have a battery of 30V connected to a resistance R of15 Ω
in a simple circuit. In our circuit we also include an ammeter to read
the current and a voltmeter to read the potential difference across the
resistance R.
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