An ideal battery with an emf of = 14.0 V is connected to a network of resistances with R1 = 6.00 , R2 = 12.0 , R3 = 4.00 , R4 = 3.00 and R5 = 5.00 . What is the potential difference across R5?
EMF = iR
Req = R + R + R in series, 1/Req = 1/R + 1/R + 1/R in parallel
The Attempt at a Solution
I have found that R1 and R2 are in parallel, which R12 is then in series with R3. R4 and R5 are in series with each other, and thus R123 and R45 are in parallel.
Finding the equivalent resistance, 1/R1 + 1/R2 = 1/R12 = 4, then R12 + R3 = 8. R4 + R5 = 8,
And 1/R123 + 1/R45 = 4 for the equivalent resistance.
Now that I have formed this circuit into a single loop with Req = 4, and EMF = 12, I can find the current, i, running through it, thus 12/4 = 3 = i
The question asks for the potential running across R5. I know that the current will be the same everywhere in the loop, so I solved for the lower branch with R4 and R5.
Knowing i=3, and Potential is V=i(r), I get 3*5 to be 15, but this is not correct.
The reason I'm trying to figure this out is because I have the feeling it will be on the next test, and I can't seem to get it.