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 Current Law**, aka **The Junction Rule**:

**Σ I _{in} = Σ I_{out}**

for any location, the total current flowing in

must equal the total current flowing out

for any loop (path returning to starting point),

the total of all voltage changes must be zero

**Wire:ΔV _{wire} = 0**

any two points connected by just wire, with nothing

else blocking the path, MUST have the same voltage!

ΔV

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

ΔV

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

ΔV

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

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 |
||

R_{tot} = R_{1} + R_{2} + R_{3} + ... |
1/R_{tot} = 1/R_{1} + 1/R_{2} + 1/R_{3} + ... |

Note: for capacitors, those two formulas are reversed:

add capacitances

add their

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.

ΔV_{res} = IR (ignore the negative for now); substitute and we get

**P _{res} = I^{2}·R**

Alternatively, we could write I = ΔV / R; substitute this and we get

More units wizardry: W = A

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=V