diff --git a/index.html b/index.html index a7bacf2..1f72465 100644 --- a/index.html +++ b/index.html @@ -19,6 +19,9 @@
By Tyler Clarke
Note: doesn't include limits of multivariable functions. +
  • + That One Physics Question +
    this took me way too long to solve

    covers some pretty simple circuitry equations diff --git a/physics-thatcircuitsquestion.html b/physics-thatcircuitsquestion.html new file mode 100644 index 0000000..8eb619b --- /dev/null +++ b/physics-thatcircuitsquestion.html @@ -0,0 +1,497 @@ + + + + + + + +- no title specified + + + + + + + + + + + + + + + + + + + + + +

    Some circuit math

    + +

    by Tyler Clarke

    +

     

    +

    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.

    + +

    We’re provided a series circuit going like,

    + + +
    The question first asks for a voltage conservation loop rule representing this circuit. Essentially, the loop rule states that for any given loop circuit, the sum of potential differences across every element must be 0. Any other value will produce perpetual motion and a violent explosion. The loop rule describing this circuit is quite simple: + + + + 1.4 + + V + + + 2 + + + + + V + thick + + + + + V + thin + + + = + 0 + + + . The potential difference across every wire segment has to add to -1.4V to counteract the battery.
    + + +
    We are then asked for a charge conservation (node rule) equation. The node rule essentially states that, across any circuit branches, the sum 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: + + + + + i + + thick + 1 + + + = + + i + + thick + 2 + + + = + + i + thin + + +
    +

     

    + +
    Next, 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 + + + + R + = + + p + ∗ + L + / + A + + + . L and A are already well-known (length and cross-sectional area) and p is the resistivity, which can be found with + + + + p + = + + 1 + / + + ( + + + n + ∗ + e + ∗ + μ + + + ) + + + +  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: + + + + R + = + + L + / + + ( + + + A + ∗ + n + ∗ + e + ∗ + μ + + + ) + + + + .
    + + +
    The electron mobility here is 0.00052 m²/(V * s), and there are 3.7 * 1028 mobile electrons per cubic meter, in all of the wires. An expression for the resistance of a thick wire is thus + + + + + R + thick + + = + + 0.28 + / + + ( + + + 1.1e-6 + ∗ + 0.00052 + ∗ + 1.602e-19 + ∗ + 3.7e28 + + + ) + + + = + 0.08258435224 + + . 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: + + + + + R + thin + + = + + 0.063 + / + + ( + + + 5.7e-8 + ∗ + 0.00052 + ∗ + 1.602e-19 + ∗ + 3.7e28 + + + ) + + + = + 0.35858995054 + + .
    + + +
    Also a pretty normal, reasonable number. Because resistance sums in a circuit, we can now apply Ohm’s law to get current: + + + + + V + = + + I + ∗ + + ( + + + + + R + thick + + ∗ + 2 + + + + + R + thin + + + + ) + + + + → + + I + = + + V + / + + ( + + + + + R + thick + + ∗ + 2 + + + + + R + thin + + + + ) + + + = + + 1.4 + / + + ( + + + + 0.08258435224 + ∗ + 2 + + + + 0.35858995054 + + + ) + + + = + 2.67298685488 + + +
    + + +
    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: + + + + i + = + + 2.67298685488 + / + 1.602e-19 + + = + 1.6685311e+19 + + . That is a much less normal number.
    + +

    + + +
    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: + + + + + E + field + + = + + p + ∗ + I + / + A + + + . Expanding resistivity, we get + + + + + E + field + + = + + I + / + + ( + + + A + ∗ + n + ∗ + e + ∗ + μ + + + ) + + + + . For the thin wire, this is just   + + + + + + E + thin + + = + + 2.67298685488 + / + + ( + + + 5.7e-8 + ∗ + 0.00052 + ∗ + 1.602e-19 + ∗ + 3.7e28 + + + ) + + + = + 15.2143845096 + + + N + / + C + + . + +  For the thick wire, + + + + + + E + thick + + = + + 2.67298685488 + / + + ( + + + 1.1e-6 + ∗ + 0.00052 + ∗ + 1.602e-19 + ∗ + 3.7e28 + + + ) + + + = + 0.78838174277 + + + N + / + C + + . + +  Easy!
    + +

    Attribution: Gemma2 running locally was quite helpful reminding me which equations to use.

    + + +