Question

Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.250 A when R1 is removed, leaving R2 connected across the battery.
(a) Find R1.
Ω
(b) Find R2.
Ω

Answer 1

Answer:

a)   R₁ = 14.1 Ω,   b)  R₂ =  19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

           V = i (R₁ + R₂)

with R₁ connected

           V = (i + 0.5) R₁

with R₂ connected

           V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

           V = i ( $\frac{V}{ i+0.5} + \frac{V}{i+0.25}$ )

           1 = i ( $\frac{1}{i+0.5} + \frac{1}{i+0.25}$ )

           1 = i ( $\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }$ ) =  $\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}$

           i² + 0.75 i + 0.125 = 2i² + 0.75 i

           i² - 0.125 = 0

           i = √0.125

           i = 0.35355 A

with the second equation we look for R1

          R₁ = $\frac{V}{i+0.5}$

          R₁ = 12 /( 0.35355 +0.5)

          R₁ = 14.1 Ω

with the third equation we look for R2

          R₂ = $\frac{V}{i+0.25}$

          R₂ =$\frac{12}{0.35355+0.25}$

          R₂ =  19.9 Ω