A Bad Ohmen from The College Board Gods (Gaelan)

Opening Questions
ObjectivesI will be able to:
Words/Formulae For the Day
Work O' The DayAlright, so we've got this little cirucit. There's some gunk in the problem about internal resistance, but we don't really care about it for now—as far as this part of the problem is concerned, a battery with an internal resistance and an ideal battery next to a resistor (like it's drawn in the circuit diagram) are the exact same thing. With that out of the way, let's give our first question a read.
The College Board Gods have spoken, and we shall obey unquestioningly. Let's start by calculating the total resistance of the circuit. We've got two resistors in parallel, so we just add them up and get our equivalent resistance: ${R}_{T}=R+r$ We'll also need Ohm's law: $I=\frac{\Delta V}{R}$ Let's plug in $\varepsilon$ in for $\Delta V$ and $R}_{T$ in for $R$. $I=\frac{\varepsilon}{{R}_{T}}$ Because our resisistors are in parallel, $I$ is the same everywhere in the circuit. So, determining the resistance ($R$) of our main resistor should be as simple as a second visit to our good friend Mr. Ohm. $V=IR$ Let's plug in our equation in for $I$: $V=\frac{\varepsilon R}{{R}_{T}}$ Note that while the R's look like they might cancel out, they don't (because one is the resistance of both resistors, and one is just that of our main one). Finally, let's plug $R}_{T$ back in so that it's just in terms of their variables: $V=\frac{\varepsilon R}{R+r}$ Alright, that's the first part done. Are the gods pleased?
Not yet, it seems. This demand seems simple enough, just some algebraic manipulation. They seem to want inverses, so taking the reciprocal of both sides seems like a good start. $\frac{1}{V}=\frac{R+r}{R\varepsilon}$ That looks about right. Let's simplify a bit by splitting up the fraction: $\frac{1}{V}=\frac{R}{R\varepsilon}+\frac{r}{R\varepsilon}$ Now, we actually can cancel out some R's, so we do: $\frac{1}{V}=\frac{1}{\varepsilon}+\frac{r}{R\varepsilon}$ Finally, because they asked us to give it in terms of $\frac{1}{R}$, so let's seperate that out so that the College Board Gods (or their lowly graders) have no excuse to mark us down. $\frac{1}{V}=\frac{1}{\varepsilon}+\frac{r}{\varepsilon}\frac{1}{R}$
Graphing? The gods must be especially angry today. Of course, we obey unquestioningly, graphing the points from the table and drawing a trendline. (Note: For technical reasons, the graph above isn't labelled. On the real text, it would need labels for full points.)
When all you've got's a TI84, everything looks like a regression problem. So let's do just that. We end up with this equation: $\frac{1}{V}=\left(0.0493\right)\frac{1}{R}+0.0817$ Hey, that looks familiar! Remember this from earlier? $\frac{1}{V}=\frac{1}{\varepsilon}+\frac{r}{\varepsilon}\frac{1}{R}$ If we match up the variables, we can figure some things out: $\frac{1}{\varepsilon}=0.0817$ $\frac{r}{\varepsilon}=0.0493$ Let's solve for $\varepsilon$: $\varepsilon =12.2\text{V}$ And plug that in to find $r$. $\frac{r}{12.2}=0.049\Omega$ $r=\left(0.0493\right)\left(12.2\right)=0.601$
Unsure if the College Board Gods will ever be satisfied, we trudge on through the darkness, starting with another look at Ohm's law. $V=IR$ Or, in this case, $I=\frac{V}{R}$ We were asked to calculate the "maximum current the battery can provide." We know that in this case the voltage we're talking about is $\varepsilon$, so let's plug that in: $I=\frac{\varepsilon}{R}$ OK, so maybe if we can maximize $I$, the College Board Gods will have mercy. $\varepsilon$ is constant, so it looks like mimimizing $R$ is our best bet. They're asking about the batery itself, so we don't care about the other resistor in the circuit. However, we still have the internal resistance in the battery itself, $r$, and we can't get any lower than that. Therefore, $I=\frac{\varepsilon}{r}=\frac{12.2}{0.601}=20.3\text{A}$
The high one. There, we're done. We've pleased the—
Darn. So this is something we haven't learned yet, but luckily it's pretty simple. A voltmeter contains a resistor with its own resistance. If the voltmeter's resistance is low, a lot of current will pass through it ($I=\frac{V}{R}$ and all that), which means that the voltmeter itself will have a fairly big impact on the ciruit, which could result in an inaccurate mesurement. Therefore, voltmeters with a high internal resistance will be more accurate. OK, now it seems that the College Board Gods are satisfied. Reference
