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<title>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?</title>
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
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
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master and sharpen your skills and techniques in comptetitive problem solving. Some of these problems were a source
of inspiration for Jaan Kalda’s handouts and to some NBPhO problems. You may find problems from old IPhO
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<meta property="og:description" content="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?">
<title>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?</title>
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>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
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>
<p>
Applying Kirchhoff Second Law (Loop's law). For loop $I$
Applying Kirchhoff Second Law (Loop's law). For loop $I$
$$-i_{11}R_1-i_{22}R_2+i_{21}R_2=0$$
$$-i_{11}R_1-i_{22}R_2+i_{21}R_2=0$$
$$-i_{11}R_1+R_2(i_{21}-i_{22})=0 \;(1)$$
$$-i_{11}R_1+R_2(i_{21}-i_{22})=0 \;(1)$$
As $i_{11}R_1 = 10 i_{12}R_1$, that is, $i_{1n} = 10 i_{1(n+1)}$, hence
As $i_{11}R_1 = 10 i_{12}R_1$, that is, $i_{1n} = 10 i_{1(n+1)}$, hence
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
