two unknown resistors a and b are connected together. when they are connected in series their combined resistance is 15 ohm. When they are connected in parallel, their combined resistance is 3.3 ohm. What are the resistances of A and B?
R1+R2 = 15 Ohms
R1*R2/(R1+R2) = 3.3 Ohms
In Eq2, replace R1+R2 with 15 Ohms:
R1*R2/(15) = 3.3
R1*R2 = 49.5
R1 = 49.5/R2
In Eq1, Replace R1 with 49.5/R2:
(49.5/R2)+R2 = 15
R2^2 + 49.5 = 15R2
R2^2 - 15R2 + 49.5 = 0
Use Quad. Formula:
R2 = 10.1 Ohms
R1 + 10.1 = 15
R1 = 4.9 Ohms
Well, well, well, what do we have here – two resistors causing all this resistance confusion? Let's solve this riddle!
Let's call the resistance of resistor A "R_A" and the resistance of resistor B "R_B".
When connected in series, the resistances add up. So, we have the equation R_A + R_B = 15 ohms.
When connected in parallel, things get a bit trickier. The formula is 1/R_parallel = 1/R_A + 1/R_B. In this case, R_parallel = 3.3 ohms.
So, let's plug in the values and see if we can tickle the funny bone of math:
1/3.3 = 1/R_A + 1/R_B.
Now, let's make things more amusing. It's time for a math magic trick! We'll solve one equation with two unknowns. Abracadabra!
R_A = 15 - R_B. (We got that from the first equation.)
Plug that into the second equation and we have:
1/3.3 = 1/(15 - R_B) + 1/R_B.
Now, let's do a little math circus to solve for R_B:
1/3.3 = (R_B + 15 - R_B) / (R_B(15 - R_B)).
1/3.3 = 15 / (R_B(15 - R_B)).
Now, multiply both sides by 3.3 to get rid of that pesky fraction:
3.3/3.3 = (15 / (R_B(15 - R_B))) × 3.3.
1 = (49.5 / (R_B(15 - R_B))).
Now, my clownly friend, let's simplify even further:
R_B(15 - R_B) = 49.5.
Expand it out:
15R_B - R_B^2 = 49.5.
Rearrange to solve for R_B:
R_B^2 - 15R_B + 49.5 = 0.
Now, let's apply some math stand-up comedy:
Using the quadratic formula, we get:
R_B = (15 ± √(15^2 - 4 × 1 × 49.5)) / (2 × 1).
R_B = (15 ± √(225 - 198)) / 2.
R_B = (15 ± √27) / 2.
With a touch of mathematical bravado, we have:
R_B = (15 ± 3√3) / 2.
So, there you have it! The resistances of A and B are:
R_A = 15 - R_B = 15 - (15 ± 3√3) / 2.
As a court jester of math, I must tell you that these are the possible values of resistances A and B. You can calculate each one separately.
Enjoy your resistor comedy show!
Let's denote the resistance of resistor A as Ra and the resistance of resistor B as Rb.
When resistors A and B are connected in series, their combined resistance is equal to the sum of their individual resistances:
Ra + Rb = 15 ohm --- (Equation 1)
When resistors A and B are connected in parallel, their combined resistance is given by the formula:
1 / (Ra + Rb) = 1 / Ra + 1 / Rb = 1 / 3.3 ohm --- (Equation 2)
To solve these equations and find the values of Ra and Rb, we can use a method called substitution.
From Equation 1, we can rearrange it to express Ra in terms of Rb:
Ra = 15 ohm - Rb
Substituting this value of Ra into Equation 2:
1 / (15 ohm - Rb) + 1 / Rb = 1 / 3.3 ohm
Now, let's calculate the values of Ra and Rb step-by-step:
1 / (15 ohm - Rb) + 1 / Rb = 1 / 3.3 ohm
Multiplying all terms by (15 ohm - Rb) * Rb * 3.3 ohm:
Rb * 3.3 ohm + (15 ohm - Rb) * 3.3 ohm = (15 ohm - Rb) * Rb
3.3 * Rb + 49.5 ohm - 3.3 * Rb = 15 * Rb - Rb^2
49.5 ohm = 15 * Rb - Rb^2
Rearranging the equation:
Rb^2 - 15 * Rb + 49.5 ohm = 0
This is a quadratic equation. Let's solve it using the quadratic formula:
Rb = [ -(-15) ± sqrt((-15)^2 - 4 * 1 * 49.5) ] / (2 * 1)
Simplifying:
Rb = [ 15 ± sqrt(225 - 198) ] / 2
Rb = [ 15 ± sqrt(27) ] / 2
Rb = [ 15 ± 5.196 ] / 2
There are two possible solutions for Rb:
1. Rb = (15 + 5.196) / 2 = 10.098 ohm
2. Rb = (15 - 5.196) / 2 = 4.902 ohm
Now that we have the possible values of Rb, we can substitute them back into Equation 1 to find the values of Ra:
1. If Rb = 10.098 ohm:
Ra = 15 ohm - Rb = 15 ohm - 10.098 ohm = 4.902 ohm
2. If Rb = 4.902 ohm:
Ra = 15 ohm - Rb = 15 ohm - 4.902 ohm = 10.098 ohm
Therefore, the possible values for the resistances of A and B are:
1. Ra = 4.902 ohm and Rb = 10.098 ohm
2. Ra = 10.098 ohm and Rb = 4.902 ohm
To determine the resistances of unknown resistors A and B, we can use a system of equations based on their behavior when connected in series and parallel.
Let's assign the resistance of A as 'x' and the resistance of B as 'y'.
When resistors A and B are connected in series, the total resistance is the sum of their individual resistances:
Resistance in series: R-series = A + B = x + y = 15 ohms -- Equation 1
When resistors A and B are connected in parallel, the reciprocal of the total resistance is the sum of the reciprocals of their individual resistances:
Reciprocal resistance in parallel: (1 / R-parallel) = (1 / A) + (1 / B) = (1 / x) + (1 / y) = 1 / (1/x + 1/y) = 3.3 ohms -- Equation 2
Now we have a system of two equations with two unknowns:
Equation 1: x + y = 15
Equation 2: 1 / x + 1 / y = 1 / 3.3
To solve this system of equations, we can rearrange Equation 2 and solve for y:
1 / y = 1 / 3.3 - 1 / x
1 / y = (x - 3.3) / (3.3x)
y = (3.3x) / (x - 3.3)
Substitute this value of 'y' into Equation 1:
x + (3.3x) / (x - 3.3) = 15
Multiply through by (x - 3.3) to eliminate the denominator:
(x - 3.3)x + 3.3x = 15(x - 3.3)
x^2 - 3.3x + 3.3x = 15x - 49.5
x^2 = 15x - 49.5
Rearrange the equation by moving all terms to one side:
x^2 - 15x + 49.5 = 0
Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For the quadratic equation x^2 - 15x + 49.5 = 0, a = 1, b = -15, and c = 49.5.
Plugging these values into the quadratic formula:
x = (-(-15) ± √((-15)^2 - 4 * 1 * 49.5)) / (2 * 1)
Simplifying further:
x = (15 ± √(225 - 198)) / 2
x = (15 ± √27) / 2
Now, we have two possible values for x. Let's calculate both:
x1 = (15 + √27) / 2 ≈ 13.563 ohms
x2 = (15 - √27) / 2 ≈ 1.437 ohms
If we rearrange Equation 1 to solve for y:
y = 15 - x
Substituting the two possible values of x back into this equation, we find the corresponding values of y:
For x1: y1 ≈ 15 - 13.563 ≈ 1.437 ohms
For x2: y2 ≈ 15 - 1.437 ≈ 13.563 ohms
Therefore, the resistances of A and B are approximately:
For A: 13.563 ohms
For B: 1.437 ohms