I apologize. Im just kinda frustrated. What kind of consequences will I deal with running a home amp with car subs?
Car speakers are designed to be used in a space that's somewhat confined. They can definitely be heard outside of the vehicle's interior, but they're tailored to produce a specific response in the car, not in a room. However, car subs can work in a house, but they usually don't sound the same.
You asked about why car speakers don't necessarily work with home audio. The main reason car speakers are usually 4 Ohm is due to the fact that a car works on a 12V DC power supply and a house uses 120V AC. That's a huge difference. a 12V system can't develop high power unless it has higher voltage to work with and unlike AC current, you can't just use a transformer to increase the voltage if the output transistors need it. The output from a head unit with common ground speaker wires is about 3 Watts/channel and when they started to make floating ground output chips, each channel was outputting 12 Watts. If you know car audio, you have seen ads stating that car power amps often use MOS-FET output transistors. Well, some of those need 100V DC for the power supply and making 100V DC from 12V DC that will drive a decent amount of current is a pretty good trick.
It will help if you knew some of the concepts of electricity- it makes finding answers easier.
Voltage- think of it as 'electrical pressure' and is similar to water pressure. At some point, you may see it called 'electrical potential' and that's because it's potential energy- it's there, ready to be used and it can be stored (in a battery, which can ONLY be Direct Current).
Current- this is electron flow measured in Amperes, which is a specific amount of electrons/second and it's helpful to think of it as being similar to water flow rate. This is what does the work and it's also what causes wires and connections to become hot. It's also the reason one amp may handle low impedance and another won't. Power is a function of voltage vs resistance vs current. P(ower) = E²/R (Voltage, squared/Resistance). Also, P(ower) = I (Current) x E (Voltage) and for the last building block, I (Current) =E/R and and each of these can be derived if the others are known
If power is to remain constant and resistance varies, the current will vary inversely. Drop the resistance to current and current will increase.