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| <title>Savchenko Solutions</title> | | <title>Savchenko Solutions</title> |
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| content="A website with solutions to physics problems from Savenko Textbook" /> | | content="A website with solutions to physics problems from Savenko Textbook" /> |
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| <header> | | <header> |
| <h2>Solutions of Savchenko Physics Textbook</h2> | | <h2>Solutions of Savchenko Physics Textbook</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <main> | | <main> |
| <article> | | <article> |
| <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> |
| | | |
| <h4>Statement</h4> | | <h4>Statement</h4> |
| <p> | | <p> |
| $8.3.15^*$ | | $8.3.15^*$ |
| 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? | | 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> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" alt="8.3.15" width="300" /> | | loading="lazy" alt="8.3.15" width="300" /> |
| <figcaption> | | <figcaption> |
| For problem 8.3.15 | | For problem 8.3.15 |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <h4>Solution</h4> | | <h4>Solution</h4> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| Let's consider the following figure | | Let's consider the following figure |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw.png" | | <img src="draw.png" |
| loading="lazy" alt="8.3.15" width="450" /> | | loading="lazy" alt="8.3.15" width="450" /> |
| <figcaption> | | <figcaption> |
| Circuit analysis | | Circuit analysis |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p class ="TxtSolutions"> | | <p class ="TxtSolutions"> |
| Applying Kirchhoff Second Law (Loop's law). For loop $I$ | | Applying Kirchhoff Second Law (Loop's law). For loop $I$ |
| </p> | | </p> |
| <center> | | <center> |
| $-i_{11}R_1-i_{22}R_2+i_{21}R_2=0$ | | $-i_{11}R_1-i_{22}R_2+i_{21}R_2=0$ |
| </center> | | </center> |
| <center> | | <center> |
| $-i_{11}R_1+R_2(i_{21}-i_{22})=0 \;(1)$ | | $-i_{11}R_1+R_2(i_{21}-i_{22})=0 \;(1)$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| As $i_{11}R_1 = 10 i_{12}R_1$, that is, $i_{1n} = 10 i_{1(n+1)}$, hence | | As $i_{11}R_1 = 10 i_{12}R_1$, that is, $i_{1n} = 10 i_{1(n+1)}$, hence |
| </p> | | </p> |
| <center> | | <center> |
| $i_{1n} = \frac{i_{11}}{10^{n-1}} \;\forall~n\geq2 \;(2)$ | | $i_{1n} = \frac{i_{11}}{10^{n-1}} \;\forall~n\geq2 \;(2)$ |
| </center> | | </center> |
| <p calss="TxtSolutions"> | | <p calss="TxtSolutions"> |
| For loop $II$ and according $(2)$ | | For loop $II$ and according $(2)$ |
| </p> | | </p> |
| <center> | | <center> |
| $-i_{N+1}(R_1+r)+i_{2N}R_2=0$ | | $-i_{N+1}(R_1+r)+i_{2N}R_2=0$ |
| </center> | | </center> |
| <center> | | <center> |
| $-\frac{i_{11}}{10^{N-1}}(R_1+r)+i_{2N}R_2=0 \;(3)$ | | $-\frac{i_{11}}{10^{N-1}}(R_1+r)+i_{2N}R_2=0 \;(3)$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| For subsequent bifurcations (applying Kirchhoff First Law) | | For subsequent bifurcations (applying Kirchhoff First Law) |
| </p> | | </p> |
| <center> | | <center> |
| $i_{11} = i_{12}+i_{22}$ | | $i_{11} = i_{12}+i_{22}$ |
| </center> | | </center> |
| <center> | | <center> |
| $i_{22} = i_{11}-i_{12}$ | | $i_{22} = i_{11}-i_{12}$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| So, | | So, |
| </p> | | </p> |
| <center> | | <center> |
| $i_{2n} = i_{1(n-1)-i_{1n}} \;\forall~n\geq2 \;(4)$ | | $i_{2n} = i_{1(n-1)-i_{1n}} \;\forall~n\geq2 \;(4)$ |
| </center> | | </center> |
| <p calss="TxtSolutions"> | | <p calss="TxtSolutions"> |
| According $(2)$, let's modify $(4)$ | | According $(2)$, let's modify $(4)$ |
| </p> | | </p> |
| <center> | | <center> |
| $i_{2n} = \frac{i_{11}}{10^{n-2}}-\frac{i_{11}}{10^{n-1}}$ | | $i_{2n} = \frac{i_{11}}{10^{n-2}}-\frac{i_{11}}{10^{n-1}}$ |
| </center> | | </center> |
| <center> | | <center> |
| $i_{2n} = \frac{9i_{11}}{10^{n-1}} \;(5)$ | | $i_{2n} = \frac{9i_{11}}{10^{n-1}} \;(5)$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| According $(5)$ into $(3)$ | | According $(5)$ into $(3)$ |
| </p> | | </p> |
| <center> | | <center> |
| $9R_2 = R_1+r \;(6)$ | | $9R_2 = R_1+r \;(6)$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| According $(5)$, from $(1)$ | | According $(5)$, from $(1)$ |
| </p> | | </p> |
| <center> | | <center> |
| $R_1 = \frac{81}{10} R_2 \;(7)$ | | $R_1 = \frac{81}{10} R_2 \;(7)$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| Finally, solving the equation system formed by $(6)$ and $(7)$, it is obtained | | Finally, solving the equation system formed by $(6)$ and $(7)$, it is obtained |
| </p> | | </p> |
| <center> | | <center> |
| $\boxed{R_1 = 9r}$ | | $\boxed{R_1 = 9r}$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| and, | | and, |
| </p> | | </p> |
| Universidad de Oriente, Cuba<br> | | Universidad de Oriente, Cuba<br> |
| </p> | | </p> |
| </article> | | </article> |
| </main> | | </main> |
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