Series Wiring Problem

runfast220

Two resistances R1 and R2, are connected in a series across a 12-V battery. The current increases by .20 A when R2 is removed, leaving R1 connected across the battery. However the current increases by just .10 A when R1 is removed, leaving R2 connected across the battery. Find(a)R1 and (b) R2.

V=12v

2. Relevant equations
R1+R2=Rs V1+V2=V

I=V/R


3. The attempt at a solution

I+20= V/R1 I+10= V/R2

I believe the two equations above would give me my answers but I'm stumped on how to go about finding the unknowns. I know V=12 but I can't figure out how find the current or resistances.

Can somebody point me in the right direction?

LowlyPion

Looks like you need 3 equations for the 3 unknowns.

V = I*R looks like it yields all 3.

(R1 + R2) = V/I

R1 = V/(I + .1)

R2 = V/(I + .2)

Then solve right?

runfast220

Using the three equations makes sense, but I can't figure out how to find the unknowns. All I know is V, and I need to know I to be able able to find R1 and R2.

my idea was to subsitute I=V/R into the equations and then try to solve for R1 and R2.
so R1= V/(V/Rs)+.1

but I don't know what Rs is either.

LowlyPion

Don't use Rs as that is by definition R1 + R2.

Otherwise, if you must, then you have 4 equations, with the 4th unknown now Rs where Rs = R1 + R2

runfast220

ok but if I plug R1 + R2 into the equation I still have too many unknowns.

R1=V/(V/R1 + R2) +.1

In that equation I still don't know how to find R1 or R2.

negitron

Disregard; I misread the problem question.

LowlyPion

You only have 3 unknowns : I, R1, R2, because V is 12.

(R1 + R2) = V/I

R1 = V/(I + .1)

R2 = V/(I + .2)

Rewriting you have

12 = I*R1 + .1R1

12 = I*R2 + .2R2

12 = I*R1 + I*R2

Adding equation 1 to equation 2 and subtracting equation 3 yields

12 = .1R1 + .2R2 or, ...

120 = R1 + 2R2

Surely it's down hill from there.

runfast220

Here is my attempt at the solution:

(R1 + R2) = V/I R1 = V/(I + .1) R2 = V/(I + .2)
SO:
(V/(I+.1)) + (V/(I+.2))=V/I

(12/(I+.1)) + (12/(I+.2))= 12/I
24I^2 + 3.6I = 12I^2 + 3.6I +.24
12I^2 = .24
I=.02 A

R1 = V/(I+.1) = 12/(.02+.1) = 100 OHMS
R2 = V/(I+.2) = 12/(.02+.2) = 54.5 OHMS

LowlyPion

i = (√2)/10

runfast220

I = .14 A

R1 = 50 ohms R2 = 35.3 ohms

Thank you for all your help.