<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Documentations on Hydrogen Notes Site</title><link>https://hydrogen.atkin.engineer/documentation/</link><description>Recent content in Documentations on Hydrogen Notes Site</description><generator>Hugo</generator><language>en-us</language><lastBuildDate>Thu, 30 Apr 2026 08:55:00 -0600</lastBuildDate><atom:link href="https://hydrogen.atkin.engineer/documentation/index.xml" rel="self" type="application/rss+xml"/><item><title>Resistor Ladders</title><link>https://hydrogen.atkin.engineer/documentation/resistor-ladders/</link><pubDate>Thu, 30 Apr 2026 08:55:00 -0600</pubDate><guid>https://hydrogen.atkin.engineer/documentation/resistor-ladders/</guid><description>&lt;p&gt;Many students are comfortable working with individual resistors, but run into confusion when those resistors are combined into networks. A common point of misunderstanding is the voltage divider: specifically, why the current through both resistors is the same even when their resistances differ.&lt;/p&gt;
&lt;p&gt;At first glance, it can seem contradictory. Using Ohm’s law, students often reason that different resistances should imply different currents, or that the voltages across the resistors must somehow be equal. The key is to recognise how series circuits behave.&lt;/p&gt;
&lt;div class="highlight"&gt;&lt;pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;-webkit-text-size-adjust:none;"&gt;&lt;code class="language-html" data-lang="html"&gt;&lt;span style="display:flex;"&gt;&lt;span&gt; Vin
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt; o
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt; |
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt; [R1]
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt; |
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt; o------ Vout
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt; |
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt; [R2]
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt; |
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt; o
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt; GND
&lt;/span&gt;&lt;/span&gt;&lt;/code&gt;&lt;/pre&gt;&lt;/div&gt;&lt;h3 id="step-1-think-of-the-total-resistance"&gt;Step 1: Think of the total resistance&lt;/h3&gt;
&lt;p&gt;The two resistors are in series, so their total resistance is:&lt;/p&gt;
&lt;p&gt;$$
R_{\text{total}} = R_1 + R_2
$$&lt;/p&gt;
&lt;h3 id="step-2-find-the-current"&gt;Step 2: Find the current&lt;/h3&gt;
&lt;p&gt;The same current flows through all elements in a series path (provided no current is drawn from the midpoint node $V_{out}$). Therefore, the current through both resistors is:&lt;/p&gt;
&lt;p&gt;$$
I = \frac{V_{in}}{R_1 + R_2}
$$&lt;/p&gt;
&lt;p&gt;This is the critical point: &lt;strong&gt;series elements share current, not voltage&lt;/strong&gt;.&lt;/p&gt;
&lt;h3 id="step-3-find-the-output-voltage"&gt;Step 3: Find the output voltage&lt;/h3&gt;
&lt;p&gt;The output voltage $V_{out}$ is simply the voltage across $R_2$. Using Ohm’s law:&lt;/p&gt;
&lt;p&gt;$$
V_{out} = I \cdot R_2
$$&lt;/p&gt;
&lt;p&gt;Substituting the expression for $I$:&lt;/p&gt;
&lt;p&gt;$$
V_{out} = \frac{V_{in}}{R_1 + R_2} \cdot R_2
$$&lt;/p&gt;
&lt;p&gt;$$
V_{out} = V_{in} \cdot \frac{R_2}{R_1 + R_2}
$$&lt;/p&gt;
&lt;h3 id="key-insight"&gt;Key insight&lt;/h3&gt;
&lt;ul&gt;
&lt;li&gt;The &lt;strong&gt;current is the same&lt;/strong&gt; through both resistors because there is only one path for charge to flow.&lt;/li&gt;
&lt;li&gt;The &lt;strong&gt;voltage divides&lt;/strong&gt; between the resistors in proportion to their resistances.&lt;/li&gt;
&lt;/ul&gt;
&lt;p&gt;A larger resistor drops more voltage, but it does &lt;em&gt;not&lt;/em&gt; carry more current in a series circuit.&lt;/p&gt;
&lt;p&gt;Once this distinction clicks—current continuity in series versus voltage division—the behaviour of resistor ladders becomes much more intuitive.&lt;/p&gt;</description></item><item><title>First Electrical Note</title><link>https://hydrogen.atkin.engineer/documentation/first-electrical-note/</link><pubDate>Mon, 27 Apr 2026 20:12:10 -0600</pubDate><guid>https://hydrogen.atkin.engineer/documentation/first-electrical-note/</guid><description>&lt;p&gt;The intent of this website is to eventually contain comprehensive notes for all the topics which I&amp;rsquo;ve covered in Electrical Engineering. That is going from Circuits 1 through to later topics in Analog Filtering and Beyond.&lt;/p&gt;
&lt;p&gt;My motivation is to have clear resources that I can use as a starting point for any students. Answer any questions that I&amp;rsquo;ve gotten multiple times.&lt;/p&gt;
&lt;h2 id="ohms-law"&gt;Ohm&amp;rsquo;s Law&lt;/h2&gt;
&lt;p&gt;Ohm&amp;rsquo;s law relates voltage, current, and resistance:&lt;/p&gt;
&lt;p&gt;$$
V = I,R
$$&lt;/p&gt;
&lt;p&gt;Where:&lt;/p&gt;
&lt;ul&gt;
&lt;li&gt;$V$ is voltage in volts (V)&lt;/li&gt;
&lt;li&gt;$I$ is current in amperes (A)&lt;/li&gt;
&lt;li&gt;$R$ is resistance in ohms (Ω)&lt;/li&gt;
&lt;/ul&gt;
&lt;p&gt;Solving for other variables:&lt;/p&gt;
&lt;p&gt;$$
I = \frac{V}{R}
\qquad\text{and}\qquad
R = \frac{V}{I}
$$&lt;/p&gt;
&lt;p&gt;Example — a 9 V battery across a 3 Ω resistor:&lt;/p&gt;
&lt;p&gt;$$
I = \frac{V}{R} = \frac{9\ \text{V}}{3\ \Omega} = 3\ \text{A}
$$&lt;/p&gt;
&lt;p&gt;Power relations (useful checks):&lt;/p&gt;
&lt;p&gt;$$
P = VI = I^2 R = \frac{V^2}{R}
$$&lt;/p&gt;</description></item><item><title>First Note</title><link>https://hydrogen.atkin.engineer/documentation/quickstart/</link><pubDate>Mon, 01 Jan 0001 00:00:00 +0000</pubDate><guid>https://hydrogen.atkin.engineer/documentation/quickstart/</guid><description>&lt;p&gt;Test Note. This will be removed&amp;hellip; Probably.&lt;/p&gt;</description></item></channel></rss>