Wednesday, September 18, 2013

3.4, the bonus section of bonus sections

Bonus section!
Section 3.4 is entitled “Electrical Circuits.”

There are quite a few things you can put into a circuit, but for the sake of this section, only four of these are important: a voltage source, an inductor, a capacitor, and a resistor.

Very briefly, a voltage source (think battery or generator) supplies the voltage to the circuit. Voltage causes electrons to move through the circuit, and the rate at which these electrons flow is called current. A resistor limits the current. A capacitor stores charge. An inductor resists the change in current that passes through it.
If you want more information about this stuff, I recommend this website, although you could always go to Wikipedia: http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Here’s what a circuit with all four of these items looks like:

Well, this is what the book portrays these things as. You would think voltage would be labeled with a V (especially when you consider that energy is denoted as an E), but sometimes voltage is referred to as the electromotive force, or emf. So in this case, a voltage source is labeled with an E. Since the voltage can sometimes be variable (changing) instead of constant through the circuit, we will denote the voltage source as a function of time, E(t).

Also, let’s talk about units. The most efficient way of doing this is to combine everything into one nice-looking chart.

Item
Denoted with
Units
Units denoted with
Voltage Source
E
Volts
V
Current
I
Amperes (Amps)
A
Resistor
R
Ohms
Ω
Inductor
L
Henrys
H
Capacitor
C
Farads
F
Charge
Q
Coulombs
C

Perhaps the next thing you’re asking is, “How do we solve a circuit such as this one?”

My answer will be, “With physics, of course! [And differential equations. Those too.]”

In order to deal with a circuit such as this one, we need to have some handy laws and rules and equations to govern ourselves with. The book calls them component laws, and we’ll look at five of them.

1. Ohm’s Law: The drop across a resistor is proportional to the current.

2. Faraday’s Law: The voltage drop across an inductor is proportional to the rate of change of current.
3. Capacitance Law: The voltage drop across a capacitor is proportional to the charge on the capacitor.
4. Kirchhoff’s voltage law (KVL): The sum of the voltage drops around any closed loop is zero.
5. Kirchhoff’s current law (KCL): The sum of currents flowing into a junction equals the sum of the currents flowing out of that junction (the book says “the sum of currents flowing into any junction is zero” (129), which is equivalent).
I gave very simple explanations for beautiful physics equations. If you care, here’s some more information on them (I especially enjoy Faraday’s Law, but that’s just me).

Capacitance Law: http://www.facstaff.bucknell.edu/mastascu/elessonsHTML/LC/Capac1.htm (It’s under the section “Voltage-Current Relationships In Capacitors”)

The beautiful thing about KVL is that means we can sum the voltage across our three items and the voltage source, and that resulting voltage will be zero.

The book has the opposite (meaning the voltage source voltage is negative and the other three are positive), and those are equivalent. When performing KVL, I tend to think of the circuit having a total positive voltage E that is coming from the voltage source (since no other item in our circuit will add any voltage into the circuit). Then I subtract the voltage that goes across each of the items in our circuit. You can do it the opposite way, but just make sure you keep track of which is negative and which is positive.
Something that helps with keeping track of signs is to sum all the voltages going across the items in your circuit and setting that equal to the total voltage coming from the voltage source.

We can rewrite the voltages going across our three items since we have handy laws for each of our three items. Thus
We defined current as the rate at which electrons (read: charge) flows. This means that I = dQ/dt and we can eliminate the Q from our equation. By differentiating each side of the equation, we get


Since this is a differential equation and this is a differential equations blog, we should set some initial conditions to make our lives easier. Usually, when we start off with a circuit, the charge on the capacitor is zero (meaning the capacitor is fully discharged) and the initial current is also zero. However, we still have a tricky second-order differential equation in our problem. We haven’t formally learned how to solve a second-order differential equation (spoiler alert: it’s going to be a thing in the future), so we’re going to save the second order-ness for a different time (perhaps our bonus section will make a brilliant comeback in the future).

If we just remove the capacitor from the circuit, then our equation becomes

Now this is something we can solve. It’s much easier when you already know the resistance and inductance of the circuit. It’s also much easier when the voltage is a constant rather than a function of time. In any event, I will leave you with this newfound knowledge of electrical circuits and bid you adieu until chapter 4.

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