Physics: Resistors and Current

Question

Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.28 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.087 A when R1 is removed, leaving R2 connected across the battery. Find (a) R1 and (b) R2.

my last problem out of my 30ish H.W. problems. Would appreciate it if anyone could help me out. Thanks

Answer 1

V=IR is all you need for this question.

In the first situation you have R1 and R2 connecting in series and resistors in series add so:

V = I (R1 + R2)   I will call this equation one

or

12 = I (R1 + R2)      

The second situation leaves out Resistor 2 and so:

V = (I + 0.28) R1     I will call this equation two

or

12 = (I + 0.28) R1      

The third situation leaves out Resistor 1 and so:

V = (I + 0.087) R2     I will call this equation three

or

12 = (I + 0.087) R2  

So now we have three equations and three unknowns:

V = I (R1 + R2)

V = (I + 0.28) R1

V = (I + 0.087) R2

Simply solve for R2 R1 or I and plug it into another one. V of course is the same for all three situations.  

Answer 2

its a system of equations type problem, R1+r2=12/ i if i recall the formula so put in the other value that r1=12/( i+.28) and similarily for r1 then equate the two sides for i and solve