Mr. 7, welcome to the forum!
E is the unknown.
I= 3
R= 1 (to keep it simple)
Therefore E= 3x1, so E=3.
Now if I raise the amps by 1 I get this setup:
E is the unknown
I= 4
R= 1
Therefore E= 4x1, so E=4.
Raising the amps in-turn raises the volts.
But I've always heard that "Raise the amps, lower the volts. Raise the volts, lower the amps"
Let me try my usual "plain ol' English" style of explanation.
Ohm's Law says 1 volt will push 1 amp through 1 ohm. If you raise the voltage or lower the resistance, the current rises; if you lower the voltage or raise the resistance, the current reduces.
As the voltage varies, the current varies proportionately. If we could vary the resistance, the current would vary proportionately, except also inversely. But, that requires starting with a new resistance in our Ohm's Law calculations.
If you apply 120 volts to a 10 ohm resistance, the current will be (120/10) 12a, and the power will be (120*12) 1440w. If you apply 240v, the current will be (240/10) 24a, and the power will be (240*24) 5760w.
Note that, because doubling the voltage doubles the current, and power is voltage times current, the power actually quadruples. That explains why it's easy to burn up equipment applying too much voltage.
How would you "raise the amps" from 3 to 4 in your example? You have to remember which parameters are the constants, and which are the variables. Which are causes, and which are effects.
The problem is that you're confusing pure math from how we use the numbers in the real world. Your example follows the rules of math, but that only explains how we apply math to our equipment.
As far as Ohm's Law is concerned, we normally consider the resistance to be a constant, the applied voltage to be our controllable variable, and the resultant current, and thus, resultant power the results.
If we're designing a supply for a load, we have to start with the desired work to be done, which we're usually given using designed supply voltage, and the resulting power in watts at that applied voltage.
Let's say the customer wants to supply a 2,400 watts electric heater, and we have to choose between using 120v or 240v to supply it. Math tells us:
2400/120 = 20a
2400/240 = 10a
Notice the constant, which is the power in this case. But, we can't use the same heating element for both voltages if we want the same power output. So, the resistance is not a constant, the desired work is.
For operation on a 120v supply, the heater's resistance would need to be (120/20) 6 ohms,, whereas for use on a 240v supply, it would need to be (240/10) 24 Ohms. That means different euipment for a voltage change.
Also, operation at the lower voltage requires a greater supply ampacity. For our electric range to have the same capacity at 120v, it would require twice the current. Would you want to install a 100a 120v circuit for a range?
The point of all this is that operation on a different supply voltage requires an equipment change, giving us new numbers to use with Ohm's Law. We normally don't simply adjust the voltage alone, unless we're varying the load.
Considering something like a stun gun makes sense of this statement, because it uses high voltage, and if raising the voltage raised the amps then a stun gun wouldn't be a stun gun, but rather a kill gun xD haha.!
A stun-gun uses a very high voltage, which would normally drive a very high current through the body, but the current output is intentionally limited, so Ohm's Law is, let's say, skewed because the source isn't unlimited.
If the body had an unusually low resistance between the contact points for some reason, such as if they happened to hit, say, a belt buckle, the voltage would drop dramatically, as if there was a resistor in series with the source.
In fact, a source who's output is considered to be of a high impedance, meaning the output voltage is easily affected by the load, is modeled as if a resistor actually was in the circuit. Using that " series resistor," Ohm's Law still applies.
