Intro to Electrical Circuits

Introduction to Electric Circuits

Electrical Terminology

Name Symbol Units Definition Conceptual Explanation
charge Q or q coulombs, C the stuff that's flowing along the wire
current I amperes, 1 A = 1 C/s I = ΔQ / Δt how much charge flows past per second
voltage V volts, 1 V = 1 J/C ΔV = ΔE / ΔQ energy density; electrical "push"; similar to pressure
EMF ε volts, V "electromotive force" (ΔV) provided by a battery
resistance R ohms, 1 Ω = 1 V / A R = | ΔV / I | "difficulty" for current to flow; also R = Δx/(A·k)
capacitance C farads, 1 F = 1 C / V C = | Q / ΔV | how "efficient" a capacitor is at storing charge
power P watts, 1 W = 1 J / s P = ΔE / Δt how quickly energy is being transferred; see below

Kirchhoff's Circuit Laws

Kirchhoff's Current Law, aka The Junction Rule:
Σ Iin = Σ Iout
for any location, the total current flowing in
must equal the total current flowing out

Kirchhoff's Voltage Law, aka The Loop Rule:
Σloop ΔV = 0
for any loop (path returning to starting point),
the total of all voltage changes must be zero

Voltage Drop Formulas

ΔVwire = 0

any two points connected by just wire, with nothing
else blocking the path, MUST have the same voltage!

ΔVbat = +ε

a battery provides a constant voltage boost no matter what;
it creates a higher voltage zone and a lower voltage zone
much like a pump with high pressure and low pressure

ΔVres = –I·R

a resistor provides a voltage drop proportional to the rate
at which current is flowing through it, with its resistance R
being the constant of proportionality... alternatively, the
voltage difference causes current to flow (from high to low),
with more voltage difference causing more current

ΔVcap = –Q / C

a capacitor provides a voltage drop proportional to the amount
of charge stored on its plates (high voltage at the positive plate,
low voltage at the negative plate), with its capacitance C being the
constant of proportionality... alternatively, the voltage difference
causes charge to be stored, with more voltage meaning more charge

Resistors in Series versus Resistors in Parallel

Series Parallel
all resistors get the same full current all resistors get the same full voltage
but split up the available voltage,
with largest R getting more voltage
but split up the available current,
with largest R getting less current
Rtot = R1 + R2 + R3 + ... 1/Rtot = 1/R1 + 1/R2 + 1/R3 + ...

Note: for capacitors, those two formulas are reversed:
add capacitances directly for capacitors in parallel;
add their inverses for capacitors in series.

A Few Notes About Power:

We define power as ΔE / Δt, the rate at which energy is being
transferred–for ANY energy, not just in electrical circuits!–but in
electrical circuits there are some very useful shortcuts. Consider the
definitions I = ΔQ / Δt and V = ΔE / ΔQ. Multiply those:
I·V = (ΔQ / Δt) · (ΔE / ΔQ) = ΔE / Δt = P, power!
This gives us a convenient new formula:
P = I·V
(that V should really be ΔV, but it usually isn't written out)
As for units, this means that watts = amps · volts.

This can be used for a battery (how much power is being provided)
or for a resistor (how much power is being dissipated). This gives us
useful information like the brightness of a light bulb, or the
loudness of a buzzer, or the work output of a motor, and so on.

Also: thanks to conservation of energy, the total power provided by
all the batteries in the circuit should match the total power
dissipated by all the resistors in the circuit.

For a resistor specifically, we can make other substitutions.
ΔVres = IR (ignore the negative for now); substitute and we get
Pres = I2·R
Alternatively, we could write I = ΔV / R; substitute this and we get
Pres = V2 / R
More units wizardry: W = A2·Ω = V2

It's useful to have all three of these formulas, because then you don't need
to know ALL of V, I, and R; you can jump straight to calculating power as
soon as you know any two of them. You can also easily compare the power of
two resistors that both have the same value of something; for instance,
resistors in parallel must have the same V, and presumably their Rs are
known, so using P=V2/R we can see that larger R means less P.