Question

Two resistors have resistances R1 and R2. When the resistors are connected in series to a 12.6-V battery, the current from the battery is 2.07 A. When the resistors are connected in parallel to the battery, the total current from the battery is 8.98 A. Determine R1 and R2. (Enter your answers from smallest to largest.)

Answer 1

When R1 = 2.193, R2 = 3.894

When R1 = 3.894, R2 = 2.193

Step-by-step explanation:

We are told that when R1 and R2 are connected in series, the voltage is 12.6 V and the current is 2.07 A.

Formula for resistance is;

R = V/I

R = 12.6/2.07

R = 6.087 ohms

Since R1 and R2 are connected in series.

Thus; R1 + R2 = 6.087 ohms

R1 = 6.087 - R2

We are also told that when they are connected in parallel, the current is 8.98 A.

Thus, R = 12/8.98

R = 1.403 ohms

Thus;

(1/R1) + (1/R2) = 1/1.403

Let's put 6.087 - R2 for R1;

(1/(6.087 - R2)) + (1/R2) = 1/1.403

Multiply through by 1.403R2(6.087 - R2) to get;

1.403R2 + 1.403(6.087 - R2) = R2(6.087 - R2)

Expanding gives;

1.403R2 + 8.54 - 1.403R2 = 6.087R2 - (R2)²

(R2)² - 6.087R2 + 8.54 = 0

Using quadratic formula, we have;

R2 = 2.193 ohms or 3.894 ohms

Thus,

R1 = 6.087 - 2.193 or R1 = 6.087 - 3.894

R1 = 3.894 or 2.193

When R1 = 2.193, R2 = 3.894

When R1 = 3.894, R2 = 2.193