Why Does Ohm's Law Calculation for External Resistor R Seem Incorrect?

[vsage]

A 12.0 V battery has an internal resistance r. The measured voltage (emf of the battery) is 12.0 V. When connected to a resistor R the terminal voltage is 11.7 V and the current is 0.10 Amperes. What is the value of the external resistor R in ohm?

Ok I thought the problem was simple but for some reason I keep getting it wrong. (E denotes electromotive force). I set up conservation of energy like this:

E - V(resistor) = 11.7

E = 12 in this case and V = i*R = 0.1*R
so..

12 - 0.1*R = 11.7

R equals 3 from my equation but it's wrong. Dead wrong. Any ideas? It's the language that seems most unclear to me.

 

[Sirus]

Try this:
$$I_{T}=\frac{V_{applied}}{R_{T}}$$
and remember what the total resistance is in this case.

 

[vsage]

Maybe my problem is more how the question is worded.. here is what I interpreted from what you said. A simple yea or nay will either send me back to the books or confirm I understood something:

It = Vapplied/Rt

since the resistors (r and R) are in parallel, Rt = R + r. Current is constant in the system so It = 0.1a but I'm a little confused as to what Vapplied is. Is that the emf? So I'm left with

0.1 = 12V / (r + R) but now I have two unknowns. ugh. Undoubtedly little r was given to me in the question but I'm having trouble wading through the semantics. I see that I haven't used that 11.7 yet.

Edit wait: emf = 0.1 * r right? So r must be 1.2..

0.1 = 12V / (1.2 + R)

R = 118.8? Edit again: Got it right. Thanks I can be so nearsighted sometimes. That was discussed in the lecture I came from not two days ago!