Solve for the equation resistance in each of the scenario: a) a 200 { , 250{ and a 300 resistor connected in series. B) same numbers as (A) but parallel
a) When resistors are connected in series, the total resistance is the sum of the individual resistances.
Therefore, the total resistance is:
R_total = R1 + R2 + R3
Given that R1 = 200Ω, R2 = 250Ω, and R3 = 300Ω, we can substitute these values into the equation:
R_total = 200Ω + 250Ω + 300Ω
R_total = 750Ω
So, the total resistance when the resistors are connected in series is 750Ω.
b) When resistors are connected in parallel, the total resistance is given by the reciprocal of the sum of the reciprocals of the individual resistances.
Therefore, the total resistance is:
1/R_total = (1/R1) + (1/R2) + (1/R3)
Rearranging the equation, we have:
1/R_total = (1/200Ω) + (1/250Ω) + (1/300Ω)
Now combine the fractions under a common denominator:
1/R_total = (15/3000) + (12/3000) + (10/3000)
1/R_total = (15 + 12 + 10)/3000
1/R_total = 37/3000
Now, take the reciprocal of both sides to solve for R_total:
R_total = 3000/37 Ω
R_total ≈ 81.08 Ω
So, the total resistance when the resistors are connected in parallel is approximately 81.08 Ω.