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<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?
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<img src="statement.png"
loading="lazy" alt="8.3.15" width="300" />
<figcaption>
For problem 8.3.15
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<h3>Solution</h3>
<p>
Let's consider the following figure
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<figure>
<img src="draw.png"
loading="lazy" alt="8.3.15" width="450" />
<figcaption>
Circuit analysis
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</figure>
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<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,
$$\boxed{R_2 = \frac{10}{9}r}$$
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<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>
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