# If batteries are a source of energy, would not lower-valued resistors cause a violation of the conservation of energy?

Here's the conundrum I have been facing. Yesterday I asked the question: how can batteries constantly motivate electrons to complete a circuit? (What maintains constant voltage in a battery?) After some thought I realized that the concept is that of an energy source. The batteries are a source of energy, and regardless of the mechanism, they will constantly 'push' for a goal. My solution finally came when I realized that speaking about someone's 'motivation' was identical to the constant push of the battery. Here's the issue:

My sister wanted to get me coffee yesterday for my birthday. She was attempting to give it to me before my class started. Unfortunately, she had difficulty with traffic, and so was able to deliver only after my class started. I realized when she told me that she tried as hard as she could, and that what really mattered was her motivation. The push she gave was constant regardless of the resistance she encountered. The parallel here to the constant push of a battery was strong; voltage didn't matter in respect to resistance for a given battery.

Now I have a different issue. The voltage drop is described as the per-capita potential energy difference of electric charges at points a and b right? The total energy would then be the amount of 'capitas' multiplied by this per capita potential energy difference. More capitas mean more total energy.

Here's the problem. If the motivation of a battery is described by voltage, then if we look at a circuit with a resistor and compare it to another circuit with an equal-length resistor but with a wider surface area, the potential difference is supposed to stay the same in both cases. However, in taking the charge from one surface to the other, more energy is required for the resistor with larger surface area! The amount of energy supplied is different in either case, yet from the same 'constant' energy source. This seems so wrong to me right now.

Yet I know that Power=Voltage*(Current)^2, so this actually makes sense in terms of the equation. I wonder what is lacking in my fundamental understanding of the potential difference in comparison to the idea of motivation/constant energy source. I'm really fundamentally misunderstanding something complex here that I don't quite have my finger put on yet. Please help clear up my view of a battery as a constant pusher to line up with this changing energy ouput a battery can give.

Happy birthday. Once upon a time, there was a king and he had a sister who also liked coffee... Or let me omit this portion of the answer.

A battery is motivating the electrons to buy a round trip ticket around the circuit by the voltage $V$. The voltage is nothing else than the energy $E$ per unit charge $Q$, i.e. $V=E/Q$; the unit 1 volt is nothing else than 1 joule of energy per 1 coulomb of charge that is transmitted between the poles. A battery with a constant voltage always increases the energy of one electron (or any fixed charge) by the same amount when it runs around.

This doesn't work indefinitely. The battery ultimately gets discharged. But the capacity of the battery is unrelated to the voltage. For example, batteries in your cell phone may have a few volts. But the capacity is often measured in mAh, like 1400 mAh in my Nokia Lumia 5200. This is a multiple of the "ampere-hour" which is a multiple of one "coulomb", a unit of charge. So the capacity tells us what is the total charge (=current times time) that may flow from one pole to the other before the battery is discharged. That's completely unrelated to the (ideally) constant voltage that tells you how much energy one electron (or one coulomb) gains when it moves from one pole to the other pole.

On one side of the battery, the electrons are sitting at a higher potential, so they prefer to slide to the other side of the battery, and the circuit is the shortest path that achieves this goal. The charge doesn't want to accumulate so inside the battery, the electrons or ions are flowing so that each piece of the matter remains neutral.

I said that voltage is energy per unit charge. A gravitational field analogy is the potential energy $mgh$ per unit mass $m$, i.e. $gh$. The height difference (multiplied by the gravitational acceleration) is motivating objects to roll down. As they are rolling down, they are "spending" their potential energy. It means that this potential energy is being converted either to kinetic energy – objects speed up when they roll down the hill – or heat (if there is some friction, the acceleration may disappear, but objects are heating up).

In the case of the voltage, the electrons are running from one pole of the battery to the other, through the circuit, and they are spending their potential energy $E=QV$, analogous to $E=mgh$ in the previous example. Such electrons are quickly reaching the maximum velocity and they no longer accelerate afterwards. So the potential energy $QV$ is mostly spent on heat. With a resistor $R$ in between, the power – energy converted to heat each second – is $$P = \frac{dE}{dt} = VI = RI^2 = \frac{U^2}{R}$$ Note that the form of the power in terms of $V,I$ is just $VI$ and not $VI^2$ as you incorrectly wrote.

If you realize that $I$, the current, is nothing else than the charge crossing a particular cross section of the wire or the resistor per unit time, $$I = \frac{dQ}{dt},$$ you may rewrite $P=VI$ as $$P = \frac{dE}{dt} = V \frac{dQ}{dt}$$ which is equivalent, after you multiply it by the small time $dt$, to $$\Delta E = V \cdot \Delta Q$$ It's the same formula as before so everything is consistent.

You may also change the resistance $R$. In the gravitational analogy, it's analogous to changing the slope of the hill from which objects roll down. As you are changing the resistance, the voltage $V$ of the (idealized) battery stays the same but the current $I$ changes so that the product expressing the voltage $V=RI$ remains constant.

So if you keep the battery and double the resistance, it's like making the slope 2 times less steep. The electrons won't accelerate so quickly. Consequently, the amount of energy that is converted to energy each second, i.e. the power $$P = \frac{dE}{dt} = \frac{V^2}{R}$$ will drop to one-half of the previous value (because $R$ is in the denominator). A less steep slope would also make the objects accelerate 2 times less quickly, and therefore the heat production would drop to 50%, too. You may double the resistance (=make the slope less steep) either by reducing the cross section of the wire to 50% of the original area, or by making the length of the wire between the poles of the battery twice longer, or some (geometric) combination of these two modifications.

• Goodness gracious good going. One thing I hadn't been realizing is that in the battery electrons formed from the chemical reaction can't be recombined with the positive ions they were separated from. That might help explain why the battery is immune to pushback from resistance. Simply well done – Andres Salas Jul 18 '14 at 18:53

I think you are confused between potential difference and energy. The energy used up can vary, if the current flowing through the resistor varies.

First of all, a battery is 'not' a constant energy source. It's a constant potential source.

Secondly, if more current flows, more energy is dissipated, even though the per capita energy is constant.

Now, to settle this idea of constant voltage, motivation, etc, let me give you something that solved similar problems for me as well in the past:

Imagine the terminals of a battery to be the plates of a capacitor. When you take one charge from the positive plate to the negative plate, some work is done in the process of transportation. Now, if you bring this charge back, some energy has to be dissipated. This is similar to the potential-kinetic energy argument. This just implies that no matter what path you choose to bring the charge back to the positive plate, the same amount of energy will be dissipated, simply because of the universality of physical laws. Hence, 'constant voltage'.