## Series Wiring Problem

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?
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 Recognitions: Homework Help 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?
 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.

Recognitions:
Homework Help

## Series Wiring Problem

 Quote by 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.
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
 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.

Recognitions:
Homework Help
 Quote by runfast220 ok but if I plug R1 + R2 into the equation I still have too many unknowns.
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.
 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

Recognitions:
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 Quote by 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
i = (√2)/10
 I = .14 A R1 = 50 ohms R2 = 35.3 ohms Thank you for all your help.