Question

Three resistors are connected in a combination of series and parallel. The combination of parallel resistors R1 and R2 is in series with resistor R3. The resistors have resistance R1=12.00Ω, R2 =13.00Ω, and R3 =14.00Ω. Sketch a picture of the situation and determine the equivalent resistance of the combination of resistors

Answer 1

Answer:In this scenario, we have two resistors (R1 and R2) connected in parallel, and the combination of these two resistors is connected in series with another resistor (R3).

To determine the equivalent resistance of the combination, we can use the following formulas:

1) For two resistors connected in parallel (R1 and R2), the formula for equivalent resistance (Rp) is:

1/Rp = 1/R1 + 1/R2

2) Once we have the equivalent resistance (Rp) of R1 and R2, the formula for the total resistance (Rt) of the combination in series with R3 is:

Rt = Rp + R3

Let's substitute the given values in the formulas:

1) For the parallel combination of R1 and R2:

1/Rp = 1/R1 + 1/R2

1/Rp = 1/12Ω + 1/13Ω

Calculating the right side:

1/Rp = 13/(12*13Ω) + 12/(12*13Ω)

1/Rp = (13 + 12)/(12*13Ω)

1/Rp = 25/(12*13Ω)

Rp = (12*13Ω)/25

Rp = 6.24Ω (approximately)

2) For the total resistance of the combination (Rt):

Rt = Rp + R3

Rt = 6.24Ω + 14.00Ω

Rt = 20.24Ω (approximately)

Therefore, the equivalent resistance of the combination of resistors is approximately 20.24Ω.

Step-by-step explanation: