Two resistances, R1 and R2, are connected in series across a 9 V battery. The current increases by 0.100 A?

Question

Two resistances, R1 and R2, are connected in series across a 9 V battery. The current increases by 0.100 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.050 A when R1 is removed, leaving R2 connected across the battery. What are the two resistances?

Answer 1

Let i=V/(R1 +R2)

i1=i+0.100 ; i1=V/R1 = i+0.100
i2=i+0.050 ; i2=V/R2 = i+0.050

i R1 + i R2 = 9
R1= 9/(i+0.100)
R2= 9/(i+0.050)

Now
9 i /(i+0.100) + 9 i/(i+0.050) = 9
i /(i+0.100) + i/(i+0.050)=1

[(i+0.100) + (i+0.050) ]i =(i+0.050)(i+0.100)

2i ^2+ .150i= i^2 +0.150 i + 0.005
i^2 = 0.005
i=+sqrt( 0.005)=0.0707A

Finally
R1= 9/(i+0.100)
R1=9/(0.0707 + 0.100)=
R1=52.72 ohms

R2=9/(i+0.050)
R2=9/(.0707 + 0.050)=
R2=74.57 ohms

Answer 2

Like all physics problems you just need to construct equations from the question. So I= 9/(R1+R2) and I1 = 9/R1 and I2 = 9/R2
Now I1 = I+0.1 and I2 = I+0.05 give you 5 equation and five unknowns so good chances to solve the equations for all the unknowns. Just eliminate the unknowns one by one

Answer 3

[9/R1]-[9/(R1+R2)]=0.a hundred and fifty 9(R1+R2) - 9(R1) = 0.a hundred and fifty(R1^2 +R1R2)...a million [9/R2]-[9/(R1+R2)]=0.0.5....2 in fact i dont have a worksheet or calculator right here yet fixing those 2 simultaneous equations could supply you R1 and R2.