Respuesta :
The values of r1 and r2 are 4 ohms and 12 ohms.
Putting 4 ohms and 12 ohms in parallel gives 3 ohms.
Putting 4 ohms and 12 ohms in series gives 16 ohms.
Putting 4 ohms and 12 ohms in parallel gives 3 ohms.
Putting 4 ohms and 12 ohms in series gives 16 ohms.
Answer:
[tex]r_{1} = 4[/tex] ω
[tex]r_{2} = 12[/tex] ω
Step-by-step explanation:
The Student is able to obtain four values of resistances using only two resistors [tex]r_{1}[/tex] and [tex]r_{2}[/tex], like that:
1. Resistance with a value of [tex]r_{1}[/tex]
2. Resistance with a value of [tex]r_{2}[/tex]
3. Resistance making series with [tex]r_{1}[/tex] and [tex]r_{2}[/tex]
4. Resistance making a parallel with [tex]r_{1}[/tex] and [tex]r_{2}[/tex]
You know that the value of resistance in series is defined by the sum of each one of them, in this case as shown below:
[tex]r_{Series} = r_{1} + r_{2}[/tex]
Now, we´ll combine in pair every value of resistance given in the statement using the formula for resistance in series and in this way to know the values for [tex]r_{1}[/tex] and [tex]r_{2}[/tex].
- [tex]r_{a}[/tex] = 3 ω + 4 ω => [tex]r_{a}[/tex] = 7 ω
- [tex]r_{b}[/tex] = 3 ω + 12 ω => [tex]r_{b}[/tex] = 15 ω
- [tex]r_{c}[/tex] = 3 ω + 16 ω => [tex]r_{c}[/tex] = 19 ω
- [tex]r_{d}[/tex] = 4 ω + 12 ω => [tex]r_{d}[/tex] = 16 ω
- [tex]r_{e}[/tex] = 12 ω + 16 ω => [tex]r_{e}[/tex] = 28 ω
If you realize, the results of resistance from what we called [tex]r_{a}[/tex], [tex]r_{b}[/tex], [tex]r_{c}[/tex] and [tex]r_{e}[/tex] not corresponding with the values in the statement, but [tex]r_{d}[/tex] does.
So you guess that [tex]r_{1} = 4[/tex] ω and [tex]r_{2} = 12[/tex] ω, but you must meet the condition, resistance in parallel be equal to 3 ω. So you use the formula for resistance in parallel:
[tex]r_{Parallel} = \frac{1}{\frac{1}{r1} + \frac{1}{r2} }[/tex] or [tex]\frac{r1*r2}{r1 + r2}[/tex]
the first way:
you replaces the values for [tex]r_{1} = 4[/tex] ω and [tex]r_{2} = 12[/tex] ω in the formula for resistance in parallel and solve the operation:
[tex]r_{parallel} = \frac{1}{\frac{1}{4} + \frac{1}{12}}[/tex]
[tex]r_{parallel} = \frac{1}{\frac{1}{3}}[/tex]
[tex]r_{parallel} = 3[/tex] ω
the second way:
[tex]r_{parallel} = \frac{4 * 12}{4 + 12}[/tex]
[tex]r_{parallel} = \frac{48}{16}[/tex]
[tex]r_{parallel} = 3[/tex] ω
and you got it, the student has getting four values of resistance by using [tex]r_{1} = 4[/tex] ω and [tex]r_{2} = 12[/tex] ω and making series and parallel.