# Application of Ohm's law / Conservation of Energy

I am taking my first Engineering Electronics class, Engineering Circuits I. I seem to be understanding it, but this one example confuses me somewhat.

The simplest manifestation of Ohm's Law (V=IR) the circuit that consists of a single loop with one load and one fixed voltage source, vaguely makes sense given how I haven't had a chemical/quantum explanation of electron actions.

However, at first was assuming that the electrons coming off a 12 Volt Battery have more potential energy than a 6 Volt battery for example, because voltage is defined as joules per columb - and a columb is a fixed number of electrons.

Then we had this problem. Mathematically, with Kirchoff's law, it makes total sense. simulate this circuit – Schematic created using CircuitLab

The problem is to determine the total voltage coming off the battery, with the internal resistance of .1 Ohms. Because the battery/alternator form a closed surface Kirchhoff's law applies and we have: $$V = 12+ [(i_b ).1 Ω]$$

But we know nothing about the potential energy of the electrons coming off of the Alternator, so this suggests that my "theory" might be wrong.

Could someone please be kind enought to explain this?

• Current from the alternator doesn't only pass through the load. – Ignacio Vazquez-Abrams Sep 4 '13 at 21:23
• I know, some current goes to the load and some goes to the battery. I assume the load somehow has a fixed current draw? – user1833028 Sep 4 '13 at 21:24
• Since it's purely resistive (in the schematic), yes. – Ignacio Vazquez-Abrams Sep 4 '13 at 21:24
• ...But if we were to say, double the voltage, and if the load was purely resistive, this would increase the current. How is the load fixed, exactly? – user1833028 Sep 4 '13 at 21:27
• It would be fixed at a new value for the new voltage. – Ignacio Vazquez-Abrams Sep 4 '13 at 21:30

## 3 Answers

From my understanding of the problem, you're making it harder on yourself by thinking about a physical manifestation...

Among the abstractions we make in circuit analysis are ideal, independent sources. In the schematic you've posted, the alternator is considered an ideal, 35A current source. This means that whatever it's connected to, it will deliver 35A no matter what. As such, a current source that is not part of a closed loop is a contradiction (because no current can flow through an open circuit).

Similarly, that 12V voltage source is an ideal voltage source, that will always have 12V across its terminals. Similar to a current source without a loop between its terminals, a voltage source with its terminals shorted is a contradiction (as anything connected between two ends of a wire in a schematic are supposed to be at the exact same potential).

With that said, it's not particularly "correct" to think of the electrons coming out of the current source to have a specific potential energy. You lose a lot of the physical world when you abstract a circuit to a simple schematic like posted.

In the schematic posted, and assuming the sources ideal, given that the load has 30A passing through it, then there are 5A passing through the battery (the 12V source and the 100m$\Omega$ internal resistance) since KCL must be satisfied. Given that the power delivered/absorbed by a device is defined to be the product of voltage and current, it should be clear that for a given current, different voltages will result in different powers. In this case, the total battery voltage is 12V + (100m$\Omega$)(5A) = 12.5V. Thus the battery is, in this case, absorbing positive power with a value of (12.5V)$*$(5A) = 62.5W.

With a 6V battery, assuming the same conditions (meaning the load is still pulling 30A and the internal resistance of the battery is identical to the 12V battery--which means the load must be different, by the way), you can see how the power absorbed by the battery would be smaller.

For further thinking... consider a simple loop consisting of a voltage source and a current source, any value for either (for instance, say 1V and 1A, respectively), and both of them connected following the passive sign convention (current source pointing from (-) to (+) on voltage source). By the definitions put into place in circuit analysis, there must be 1A flowing through the loop, so the 1V source is absorbing 1W of power, while the 1A source is delivering 1A of power. Change the 1V source to, say, a 5kV source, and the power absorbed/delivered will be greater by a factor of 5000. Note that the electrons coming out of the current source now must have 5000 times the potential (by definition, since the voltage source has 5000 times the potential across it), but the current source hasn't changed. Naturally, if the 1A current source were not ideal, it may not be able to supply 1A through a 5kV battery, but in circuit analysis, these ideal sources are defined to deliver their rated spec, regardless of the circuit they are a part of (short of contradictions).

• I agree: you can have either an electron-based physical model or ideal current/voltage sources, but not both. – pjc50 Sep 5 '13 at 9:44

I think your confusion might be coming from this statement:

voltage is defined as joules per columb - and a columb is a fixed number of electrons.

In this case the number of Coulumbs flowing through the conductor (current = Coulumbs/second) is being set by the alternator and load.

Not sure if this answers your question though...

We simply have to take the schematic at face value. The alternator is modeled as an ideal current source, connected to points a and b. There is some concrete voltage across a and b for which we can solve. That voltage tells us, if we are interested, the potential energy that electrons lose as they cycle out of the alternator and back in.

Points a and b, as you know, happen to be the battery terminals, since the 0.1 ohm resistance is internal.

So the potential across the alternator is the same as across the battery terminals. Points a and b are nodes in the circuit, and we look at voltages between nodes without regard for what devices are connected to them. For any given electron at point b, it doesn't matter whether it was injected there by the battery or by the alternator.