I may have asked this question before, but I can't find it. I apologize if so.
Instead of increasing voltage, why can't you increase amps to get the same power when calculating Ohms?
Example (rounded):
8 Ohms
3.5 Amps
28 Volts
=100 watts
I can't lower volts and increase amps to get the same power.
I'm using this calculator:
https://www.rapidtables.com/calc/electric/watt-volt-amp-calculator.html (https://www.rapidtables.com/calc/electric/watt-volt-amp-calculator.html)
Because the voltage across a resistor and the current through it are bound together by Ohm's Law (https://en.wikipedia.org/wiki/Ohm%27s_law):
V = I × R
With V in volts, I in ampere and R in ohms.
So, if you have 28 V across an 8 Ω resistor the current through it will be 28/8 = 3.5 A no matter what, and vice versa.
In fact, two common ways to calculate the power dissipated by a resistor are:
P = V × I = V × (V/R) = V2 / R
and
P = V × I = (I × R) × I = I2 × R
Since the fundamental constraint is V = IxR, you can only specify two and solve for the third.
In general, with a non-linear load, the equation is V = f(I) and you should graph the function (such as for a semiconductor device).
For AC, you can use a variable transformer to obtain the desired power in a specified resistor, or to maintain the desired power when you change the resistor, from a given voltage source; this allows you to change the voltage across the load to obtain your desired power.
Sometimes a slightly different explanation helps.
You can't adjust voltage and current independently. You can reduce volts and increase current to get the same power. But to do that you have to reduce resistance.
Your question may come from an audio problem. My home stereo puts out 240 W into an 8 ohm impedance speaker. I want my car stereo with only 12 volts available to also put out 240 W. Obviously that will require 20 Amps of current. And to get that 20 Amps the speaker impedance would have to be down around 0.6 ohms. (You can work back and find that the home stereo must have a roughly 44 volt supply).
You can make these results happen in the referenced calculator by resetting between calculations. Its internal logic isn't clever enough to deal with serial changes in values.
As I said in a reply to one of your other posts, a transformer can be used to couple a speaker impedance to an amplifier that delivers its maximum power into a different impedance, but solid-state amplifiers normally avoid transformers and are designed to drive a "normal" impedance (like 4 or 8 ohms) directly. (Designing a good transformer for the 10-octave range required for high-fidelity audio is a difficult problem, and good audio transformers are heavy and expensive.)
As part of that design, you can select the DC voltage applied to the amplifier, possibly using a switch-mode power supply to obtain the desired voltage from a 12 V car system, or the 170 V DC available after rectifying the 120 V AC line. In the line-driven case, the transformer required to isolate the amplifier from the mains runs at a much higher frequency than 60 Hz, and can be far lighter.
That's how car audio amplifiers work. They have a switching power supply that converts 12V into + and - rails of sufficient voltage to push the rated power through the impedance of the speakers.
Would the continuous power still be limited by the power source's voltage and amperage? Our other discussion was about class D amp chips and the continuous ratings in power you see on Amazon with these class D chips in home audio systems powered by DC. I mean the maximum wattage a 24V 5A power supply can deliver is 120 watts and no resistance.
One of them states the class D 3116D Texas Instruments can put out 200 watts total "RMS" out of a 24V 5A power supply. I'm calling bullshit.
Would the continuous power still be limited by the power source's voltage and amperage? Our other discussion was about class D amp chips and the continuous ratings in power you see on Amazon with these class D chips in home audio systems powered by DC. I mean the maximum wattage a 24V 5A power supply can deliver is 120 watts and no resistance.
One of them states the class D 3116D Texas Instruments can put out 200 watts total "RMS" out of a 24V 5A power supply. I'm calling bullshit.
Yes you can't get something for nothing. The RMS wattage the amp delivers cannot exceed the power that it draws from the power supply.
| @ 8 ohms: 72 watts total | 8 Ohms across 24V gives 3A. 3A@24V gives 72W |
| @ 4 ohms: 100 watts total | 4 Ohms across 24V gives 6A - but the supply can only give 5A Power is: I2.R = 5*5*4 which gives 100W (V=I.R which is 20V - a 4V voltage sag) |
| @ 2 ohms: 50 watts total <----What? | 2 Ohms across 24V gives 12A - not going to happen. The supply can only give 5A. Power is: I2.R = 5*5*2 which gives 50W (V=I.R which is 10V - a 14V voltage sag) |
And I^2 R losses in the wiring (including speaker leads & connectors) comes into play.