Q: I see alot of electrical products giving its recommended input voltage for its powering up, which is fine n dandy, but almost never do they give you the recommended ampers ? Why is it so ?
A: if you know the working voltage, you know how much amps pass trough. Then it depends if the object powered up is "linear" on not linear, but assume it linear(a resistor) and powered it up in DC, you have a 10 ohm resistor and you power it up at 20 V, you get 2A.
Q: Doesnt the 2 of those give you the power (watts) for it ? What if you buy yourself an adapter and overshoot or undershoot with the power, and you get excessive heat or it not working at all ?
A: Depends. P=V*I only in DC circuits. In AC circuits, you have to care about the power factor(which is between 0 and 1). If you buy an adapter and overshot/undershot the equipment you get less or higher value of current, and precisely, if you overshot the circuit and you get some high value of current you will break the product: for instance, if you have a DC motor and it works whit 3.3V and you power it up to 9 V, you will see the motor working, and rotating very fast, then it will start to heat up and then you get a short circuit because of the heat melting the insulation between the wires.... The problem in cables, products, motors and all things like that is the current: current produce heat, and heat is not so good for the insulation.... If you undershot the equipment you get the object working, In some cases won't work, but if you have a resistor and its powered it up at 10 V or 40 V and nominally works at 100V the current will change of course, but it is still working like a resistor... but it's like having a Ferrari and drive to 5 mile/h...
Q: My next main question would be, does the ampere value drop in a circuit when passing resistors ? I know for sure voltage does. We can just calculate the total resistor value and find the amperes with the voltage source value ? Where exactly is the calculated voltage at on the resistor ? At the resistor start, the middle, the end, or just barely past it ?
A: If we measure voltage, we measure ti between two points: it's called potential difference between two points. At the same potential we have 0V (for remember it just think: if you make a step are you dead? no, so you are at 0V). so, if you have a drop it between the resistor. You should know the ohm law: V=R*I dimensionally speaking [ Voltage ] = [ohms ] *[ amps] . This formula can be always reversed, so if you know the voltage and the current, you know the resistance (between the current path)
Q:From mains wiring, what exactly dictates how many amps were getting from it ? Voltage sources come with a fixed voltage value, do they also come with a fixed amperage value ? How do we calculate them ?
A: No: in fact you can plug to your mains a load witch is 24K VA and for a certain time it will work, then what happen? the fuses will open the circuit: you request too much power and before you house start to burn it's better open the circuit.... But it's better you let go the mains: it's better for you starting to know basics principles of the electricity/electronics instead complicating your way..... i told this to you for experience.
Q: Linear power supplies often have an amps setting in them aswell, how exactly is the value being controlled ? Was i wrong with the ampers not dropping in a circuit ? If I was wrong then looking back at question number 2, how do we calculate the ampers on each resistor in a circuit ?
A: As I know, you control what's on the LCD. If you have a circuit which in the paper is a resistor of 8 ohm and you power it up whit 8V, you get 1A. If you get 10A or the max current something is wrong. Depends how much resistor you have: remember la large number of same resistor in series = an open circuit (0A), and a large number of same resistor in parallel = larger current. so, if you have a circuit whit a larger number of resistor in parallel, you get that. For calculating the drop on resistor or the current value on one of them you should see the series/ parallel formulas, + the voltage/current divider.
Hope I answered.