Answer:
the answers, the correct one is D, Rb₂ = 29.97 ohm
Explanation:
For this exercise we use ohm's law and the equivalent resistance ratio for series and parallel circuits.
Serial circuit
(Ra + Rb) is = V
(Ra + Rb) 0.111 = 10
(Ra + Rb) = 10 / 0.111 = 90.09
parallel circuit
R =
\frac{Ra \ Rb}{Ra + Rb} i_p = V
\frac{Ra \ Rb}{Ra + Rb} 0.5 = 10
\frac{Ra \ Rb}{Ra + Rb} = 10 / 0.5 = 20
we write and solve our system of equations
Ra + Rb = 90.09
\frac{Ra \ Rb}{Ra + Rb} = 20
we solve for Ra in the first equation
Ra = 90.09 - Rb
RaRb = 20 (Ra + Rb)
we substitute Ra in the second equation
(90.09-Rb) Rb = 20 [(90.09-Rb) + Rb]
90.09 Rb - Rb² = 20 90.09
Rb² - 90.09 Rb + 1801.8 = 0
we solve the quadratic equation
Rb = [90.09 ± ] / 2
Rb = [90.09 ± 30.15] / 2
Rb₁ = 60.12 ohm
Rb₂ = 29.97 ohm
the smallest value is Rb = 30 ohm
When checking the answers, the correct one is D