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?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#8.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <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>

    <h3>Solution</h3>
    <p>
      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>
      Applying Kirchhoff Second Law (Loop's law). For loop $I$
      $$-i_{11}R_1-i_{22}R_2+i_{21}R_2=0$$
      $$-i_{11}R_1+R_2(i_{21}-i_{22})=0 \\;(1)$$
      As $i_{11}R_1 = 10 i_{12}R_1$, that is, $i_{1n} = 10 i_{1(n+1)}$, hence 
      $$i_{1n} = \frac{i_{11}}{10^{n-1}} \\;\forall\~n\geq2 \\;(2)$$
      For loop $II$ and according $(2)$
      $$-i_{N+1}(R_1+r)+i_{2N}R_2=0$$
      $$-\frac{i_{11}}{10^{N-1}}(R_1+r)+i_{2N}R_2=0 \\;(3)$$
      For subsequent bifurcations (applying Kirchhoff First Law)
      $$i_{11} = i_{12}+i_{22}$$
      $$i_{22} = i_{11}-i_{12}$$
      So,
      $$i_{2n} = i_{1(n-1)}-i_{1n} \\;\forall\~n\geq2 \\;(4)$$
      According $(2)$, let's modify $(4)$
      $$i_{2n} = \frac{i_{11}}{10^{n-2}}-\frac{i_{11}}{10^{n-1}}$$
      $$i_{2n} = \frac{9i_{11}}{10^{n-1}} \\;(5)$$
      According $(5)$ into $(3)$
      $$9R_2 = R_1+r \\;(6)$$
      According $(5)$, from $(1)$
      $$R_1 = \frac{81}{10} R_2 \\;(7)$$
      Finally, solving the equation system formed by $(6)$ and $(7)$, it is obtained
      $$\boxed{R_1 = 9r}$$
      and,
      $$\boxed{R_2 = \frac{10}{9}r}$$
    </p>

    <p 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>



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