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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.
NEDS7

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.  

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