Added Luis's English solution of 8.3.15

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+ <title>Savchenko Solutions</title>
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+ <meta property="og:title" content="Savchenko Solutions"/>
+ <meta name="description"
+ content="A website with solutions to physics problems from Savenko Textbook" />
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+
+<body id="top">
+ <header>
+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <main>
+ <article>
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
+
+ <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>
+ <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>
+ </main>
+
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