Two resistors have resistances R1 and R2. When the resistors are connected in series to a 12.0-V battery, the current from the battery is 2.00 A.

Question

Two resistors have resistances R1 and R2. When the resistors are

connected in series to a 12.0-V battery, the current from the battery is

2.00 A. When the resistors are connected in parallel to the battery, the

total current from the battery is 9.00 A. Determine R1 and R2.

Please explain each step

Answer 1

Basically, this is a simple question, All you need to know is: Ohm's Law; Series and parallel formula and solving quadratic eq. Pl let me know if any step is not clear.

According to Ohm's Law V = I (amp) X R (Ω)

∴ Effective R in series = V/I = 12/2 = 6

∴ R₁ + R₂ = 6 Ω ------eq A

Or R2 = 6-R₁ ------eq B

In parallels, 1/R₁ + 1/R₂ = 12/9 = 4/3 Ω

Or (R₁ R₂ )/(R₁ + R₂ ) = 4/3 -----eq C

Solving eq A and C together, (R₁ R₂ )/(6 ) = 4/3

∴ (R₁ R₂ ) = 8, Substitute R2 from eq B, we get

(R₁) (6-R₁) = 8 OR 6R₁ – (R₁)² – 8 = 0 This can be written as

(R₁)² – 6R₁ + 8 = 0 Solve for quadratic Eq x = –b ± [√b² - 4ac]/2a

we get R1 = 4Ω then R2 = 2Ω

OR R1 = 2Ω then R2 = 4Ω