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<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>
+ <h3 id="back-link"><a href="../">$\leftarrow$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="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>
+ </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>
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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>
+ </p>
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