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<p class="paragraph-P1">Some circuit math</p>
<p class="paragraph-P2">by Tyler Clarke</p>
<p class="paragraph-P3"> </p>
<p class="paragraph-P3">I was given a pretty interesting problem in Physics 2, and figured I oughta share my solutions for all the enterprising students trying to learn about circuits.</p>
<p class="paragraph-P3">We’re provided a series circuit going like,</p>
<ul><li><p class="P4" style="margin-left:0cm;"><span class="Bullet_20_Symbols" style="display:block;float:left;min-width:0.635cm;">•</span>A battery, pushing 1.4 volts<span class="odfLiEnd"/> </p></li><li><p class="P4" style="margin-left:0cm;"><span class="Bullet_20_Symbols" style="display:block;float:left;min-width:0.635cm;">•</span>A 0.28m long thick wire with a cross sectional area of 1.1 * 10<span class="text-T1">-6 </span><span class="text-T2">m²</span><span class="odfLiEnd"/> </p></li><li><p class="P4" style="margin-left:0cm;"><span class="Bullet_20_Symbols" style="display:block;float:left;min-width:0.635cm;">•</span>A 0.063m long thin wire with cross sectional area 5.7 * 10<span class="text-T1">-8</span><span class="text-T2"> m</span><span class="text-T1">2</span><span class="odfLiEnd"/> </p></li><li><p class="P4" style="margin-left:0cm;"><span class="Bullet_20_Symbols" style="display:block;float:left;min-width:0.635cm;">•</span>Another thick wire identical to the first, going back to the battery<span class="odfLiEnd"/> </p></li></ul>
<!--Next 'div' was a 'text:p'.-->
<div class="paragraph-P3">The question first asks for a voltage conservation <span class="text-T3">loop rule</span><span class="text-T4"> representing this circuit. Essentially, the loop rule states that for any given loop circuit, the sum of potential differences across every element </span><span class="text-T3">must</span><span class="text-T4"> be 0. Any other value will produce perpetual motion and a violent explosion. The loop rule describing </span><span class="text-T3">this</span><span class="text-T4"> circuit is quite simple: </span>
<!--Next 'span' is a draw:frame. -->
<span id="Object1"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline">
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</math> </span><span class="text-T4">. The potential difference across every wire segment has to add to -1.4V to counteract the battery.</span></div>
<!--Next 'div' was a 'text:p'.-->
<div class="paragraph-P3"><span class="text-T4">We are then asked for a charge conservation (node rule) equation. The node rule essentially states that, across any circuit branches, the </span><span class="text-T3">sum </span><span class="text-T4">of the current on both branches is equal to the current flowing into the branch. Since this is a series circuit, the rules for parallel circuits don’t apply, so we can just use the standard wisdom that current must be the same at every point in a series circuit: </span>
<!--Next 'span' is a draw:frame. -->
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</math> </span></div>
<p class="paragraph-P5"> </p>
<!--Next 'div' was a 'text:p'.-->
<div class="paragraph-P6"><span class="text-T5">Next</span>, we’re asked to find the electron current in the middle of the thin wire. This is a hard one. The obvious-ish way I can see is to calculate the total resistance (electron mobility and supply is given, and we know wire dimensions, so this shouldn’t be too hard) and use Ohm’s law. Gemma2 running locally seems to agree, and helpfully informed me that the resistance is just
<!--Next 'span' is a draw:frame. -->
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<mi>A</mi>
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</math> </span>. L and A are already well-known (length and cross-sectional area) and p is the resistivity, which can be found with
<!--Next 'span' is a draw:frame. -->
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</mrow>
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</math> </span> where n is the electron density, e is the charge of an electron, and μ is the electron mobility. This combines nicely to the behemoth:
<!--Next 'span' is a draw:frame. -->
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</math> </span><span class="text-T6">.</span></div>
<!--Next 'div' was a 'text:p'.-->
<div class="paragraph-P7"><span class="text-T6">T</span><span class="text-T4">he electron mobility here is 0.00052 m²/(V * s), and there are 3.7 * 10</span><span class="text-T7">28</span><span class="text-T8"> mobile electrons per cubic meter, in all of the wires. An expression for the resistance of a thick wire is thus </span>
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<mn>1.1e-6</mn>
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</math> </span><span class="text-T10">. </span><span class="text-T8">That’s a… surprisingly normal number! I was half expecting it to be something ridiculous. 0.0826 ohms seems about right for the wires in this configuration. Now for the thin wire: </span>
<!--Next 'span' is a draw:frame. -->
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<mn>0.35858995054</mn>
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</math> </span><span class="text-T8">.</span></div>
<!--Next 'div' was a 'text:p'.-->
<div class="paragraph-P7"><span class="text-T8">Also a pretty normal, reasonable number. Because resistance sums in a circuit, we can now apply Ohm’s law to get current: </span>
<!--Next 'span' is a draw:frame. -->
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<mn>2.67298685488</mn>
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</math> </span></div>
<!--Next 'div' was a 'text:p'.-->
<div class="paragraph-P7"><span class="text-T8">Nice. A pretty normal number! Unfortunately, this is not electron current. This is amperage in coulombs per second, but we need electrons per second. Fortunately, because there are 1.602e-19 coulombs per electron, we can just divide: </span>
<!--Next 'span' is a draw:frame. -->
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</math> </span><span class="text-T8">. That is a much less normal number.</span></div>
<p class="paragraph-P7"><span class="text-T8"/></p>
<!--Next 'div' was a 'text:p'.-->
<div class="paragraph-P8"><span class="text-T11">Finally, we’re asked to find the electric field at the middle of the thin wire and at the middle of a thick wire (it is the same in both thick wires). The electric field strength in a wire is just resistivity times current density, and current density is current divided by area, so we have a simple formula: </span>
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</math> </span><span class="text-T11">. Expanding resistivity, we get </span>
<!--Next 'span' is a draw:frame. -->
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</math> </span><span class="text-T11">. For the thin wire, this is just  </span>
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</math> </span><span class="text-T11"> </span><span class="text-T12">For the thick wire, </span>
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</math> </span> <span class="text-T13">Easy!</span></div>
<p class="paragraph-P9"><span class="text-T14">Attribution: Gemma2 running locally was quite helpful reminding me which equations to use.</span></p>
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