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

a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω

Step-by-step 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 (
`(V)/( i+0.5) + (V)/(i+0.25)` )

1 = i (
`(1)/(i+0.5) + (1)/(i+0.25)` )

1 = i (
`(i+0.5+i+0.25)/((i+0.5) \ ( i+0.25) )` ) =
`(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 =
`(V)/(i+0.5)`

R₁ = 12 /( 0.35355 +0.5)

R₁ = 14.1 Ω

with the third equation we look for R2

R₂ =
`(V)/(i+0.25)`

R₂ =
`(12)/(0.35355+0.25)`

R₂ = 19.9 Ω