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  <h4>Statement</h4>
  <p>
    $8.3.15^*$
    The attenuator is a voltage divider, the circuit of which is shown in the figure. What should be the resistances $R_1$ and $R_2$ so that at each subsequent resistance $R_1$  the voltage is ten times less than at the previous one?
  </p>
  <center>
    <figure>
      <img src="statement.png"
        loading="lazy" alt="8.3.15" width="300" />
      <figcaption>
        For problem 8.3.15
      </figcaption>
    </figure>
  </center>
  <h4>Solution</h4>
  <p class="TxtSolutions">
    Let's consider the following figure
  </p>
  <center>
    <figure>
      <img src="draw.png"
        loading="lazy" alt="8.3.15" width="450" />
      <figcaption>
        Circuit analysis
      </figcaption>
    </figure>
  </center>
  <p class ="TxtSolutions">
    Applying Kirchhoff Second Law (Loop's law). For loop $I$
  </p>
  <center>
    $-i_{11}R_1-i_{22}R_2+i_{21}R_2=0$
  </center>
  <center>
    $-i_{11}R_1+R_2(i_{21}-i_{22})=0 \\;(1)$
  </center>
  <p class="TxtSolutions">
    As $i_{11}R_1 = 10 i_{12}R_1$, that is, $i_{1n} = 10 i_{1(n+1)}$, hence 
  </p>
  <center>
    $i_{1n} = \frac{i_{11}}{10^{n-1}} \\;\forall\~n\geq2 \\;(2)$
  </center>
  <p calss="TxtSolutions">
    For loop $II$ and according $(2)$
  </p>
  <center>
    $-i_{N+1}(R_1+r)+i_{2N}R_2=0$
  </center>
  <center>
    $-\frac{i_{11}}{10^{N-1}}(R_1+r)+i_{2N}R_2=0 \\;(3)$
  </center>
  <p class="TxtSolutions">
    For subsequent bifurcations (applying Kirchhoff First Law)
  </p>
  <center>
    $i_{11} = i_{12}+i_{22}$
  </center>
  <center>
    $i_{22} = i_{11}-i_{12}$
  </center>
  <p class="TxtSolutions">
    So,
  </p>
  <center>
    $i_{2n} = i_{1(n-1)}-i_{1n} \\;\forall\~n\geq2 \\;(4)$
  </center>
  <p calss="TxtSolutions">
    According $(2)$, let's modify $(4)$
  </p>
  <center>
    $i_{2n} = \frac{i_{11}}{10^{n-2}}-\frac{i_{11}}{10^{n-1}}$
  </center>
  <center>
    $i_{2n} = \frac{9i_{11}}{10^{n-1}} \\;(5)$
  </center>
  <p class="TxtSolutions">
    According $(5)$ into $(3)$
  </p>
  <center>
    $9R_2 = R_1+r \\;(6)$
  </center>
  <p class="TxtSolutions">
    According $(5)$, from $(1)$
  </p>
  <center>
    $R_1 = \frac{81}{10} R_2 \\;(7)$
  </center>
  <p class="TxtSolutions">
    Finally, solving the equation system formed by $(6)$ and $(7)$, it is obtained
  </p>
  <center>
    $\boxed{R_1 = 9r}$
  </center>
  <p class="TxtSolutions">
    and,
  </p>
  <center>
    $\boxed{R_2 = \frac{10}{9}r}$
  </center>
    <p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;">   
        BSc. Luis Daniel Fernández Quintana<br>
        Physics Department (FCNE)<br>
        Universidad de Oriente, Cuba<br>
    </p>
</article>