You have 2 resistors of unknown values you label Ra and Rb. You have an old battery and a multimeter you bought years ago for 7$ at Harbor Freight. Using the meter in voltage mode, you measure 10 V across the battery. You then connect the 2 resistors in series across the battery and use the meter in current mode to find the current flowing through the circuit. It reads 0.111A. You then connect the 2 resistors in parallel across the battery and use the meter again to measure the current now coming from the battery to be 0.5A. With this information you have gathered, you find the value of the 2 resistors. Value of smallest resistance in ohms.
a. 60
b. 90
c. 20
d. 30

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

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary