(E/e) = (R+r)/r

We each got different answers. Please assist. Solve for r.

(E/e) = (R+r)/r

Is this homework?

r = (R+r)/(E/e)

It is, but not mine. She’s doing homework, asked me to check. Now I’m not sure of either of our answers, and need a third opinion. She won’t let me tell her the right answer, only whether she is right or not, and a guide in the right direction.

r = R / (E/e - 1)

Breaking it down:

E/e = (R+r)/r
E/e = R/r + 1
E/e - 1 = R/r
r * (E/e - 1) = R
r = R / (E/e - 1)

You could go one step further and get:

r = (R*e)/(E-e)

(R+r)/r = E/e
R/r + r/r = E/e
R/r = E/e - r/r
R/r =(E/e -1)
R= r(E/e -1)
r = R/(E/e-1)
Thats what I got.

What are your two answers? It’s possible they are both correct.

In fact, I got:
r = R / (E/e - 1)

and she got:
r = (R*e)/(E-e)

So. Well.

Yup, as you presumably now realize, those are both correct (and equivalent to each other; just scaling up the numerator and denominator by a factor of e to go from one to the other)

That’s true as well, but perhaps less helpful as a “solution”, seeing as how there’s still an r on the right hand side.

Never mind.

I’m doing something wrong, I keep getting E-I-E-I-O.

Oh, shit. Sorry. :smack:

Depending on the intended application, it might be necessary to specify in the second form ( R*e/(E-e) ) that e is not equal to 0. This maybe makes the other form preferable.

It is probably implicit in the original problem that both e and r are non-zero, as it involves division by them. More importantly, the given solutions only work when E is not equal to e; if E is equal to e, then the problem is insoluble if R is non-zero, and is satisfied by all r if R is zero.

I my math classes, we were told not to leave a fraction in the denominator if we could at all help it.

Eh, the two are both valid expressions meaning (essentially) the same thing; go with whichever strikes your own aesthetic senses as best. People teach an awful lot of weird superstitions about what forms to put things in in math class (e.g., I’m constantly having to remind students “You know, there’s nothing wrong with having a square root in a denominator…”, but still, they keep twitching at it). Yeah, in many cases, you can move fractions out of the denominator in such a way as makes things notably simpler, but evaluating simplicity is always a judgement call.

It’s math class, not english class. They’re both 100% right and equivalent. Any argument otherwise is a question of style, not substance.

I suspect that grade-school math teachers — who have to check all that homework, remember, after it’s turned in — would prefer their students’ answers look more or less the same. If there are two or three or nine valid expressions for the same solution, that’s more wear and tear on the teacher’s brain when he’s grading.

But I’ve never asked a math teacher about it. That’s just my conjecture.

Let me guess… a circuit with a resistor R, and a cell with internal resistance r? Amirite?

I’ve certainly graded enough to appreciate that, but on the other hand, when a student gets an answer in a different form from the key and everyone else, you can be confident that they did it themselves.