Hi folks!

I’m trying to understand some of the fundamentals of electricity and I’m struggling with the concept of voltage drop, load, especially as it relates to Ohm’s law.

1) **How does a load draw amperage?** More specifically, how does it draw _just_ the amperage it needs? Let’s say for instance, we’re talking bout residential mains, in which we have a 12A (15A nominal) circuit with a wall outlet. At that wall outlet is ~117V (depending on the length of the wire that feeds) it, and _up to_ 12A of available current. Now, I have two loads that I can attach, let’s say, a high powered hair dryer that wants to suck 10A, and some sort of charger that maybe wants 1A.

What limits the charger to only use 1A, is it just the resistance built into the circuit? What about the hair dryer? Is it the total resistance of the device? Is this just a simple application of Ohm’s law?

We would be dealing with impedance rather than simple resistance since this is an AC circuit. But, in the end, it's the voltage and impedance that determine how much current the device will draw. You're solving I=V/Z because both V and Z are constants (for this discussion).

2) **Do all loads cause a voltage drop?** How do we know what the voltage drop for any given component is? Is it just published with the component? Does Ohm’s law have anything to do with voltage drop? Is it somehow predictable/computable/calculatable?

Appliances often specify the current draw. This includes all the complications of impedance which brings in complex numbers or vectors. Just take their word for it, they measured it with an AC ammeter. Of course Ohm's law applies! It's a LAW, not a suggestion! The voltage drop is calculated from V = I / Z. If you have the parameters for Z, yes, you can compute the answer. Not necessarily an easy calculation but definitely doable in the first semester of electric circuits.

Sometimes, appliances just give wattage. The get amps, divide wattage by voltage. A = P / V then Z = V / A. At some point in the algebra, P = I

^{2} * R.

2a) I understand Ohm’s law states that resistance will drop the amperage through a circuit by a proportionate amount, but a voltage divider works because there’s a voltage drop at the end of each resistor. How is that calculated? How does that relate to Ohm’s law?

Thanks!!!

Resistance doesn't 'drop' amperage, it drops a voltage based on the current passing through. It will, of course, limit the current.

For the voltage divider, add the two resistors (R1 + R2) and call that R

_{T}. Figure the total current I = V / R

_{T} then calculate the output voltage (across the bottom resistor, R2) as V = I * R2

ETA: If you were to add a parallel load resistor to R2 to simulate some other circuitry, the value of R2 is the parallel combination of these resistors. It would necessarily reduce the output voltage. That's why voltage dividers are not an ideal way to generate a specific voltage unless you have some kind of voltage follower (op amp) to drive the load.

ETA: Stay away from AC circuits until you have nailed Kirchoff, Thevenin and Norton for resistive circuits. The math for AC circuits get truly ugly.