More resistance is less load??

Good responses by the others above, except that the reference to E/IR should be E = IR (meaning E equals I multiplied by R). (E is commonly used in the context of Ohm’s Law to denote volts, and means the same thing in that context as V).

So when Ghosthouse referred to V/I = R, he was correct. By simple algebra V/I = R is equivalent to V (or E) = I x R = IR.

Regarding the voltages that are typically provided to speakers, for a resistive load:

Power = (I squared) x R = (E squared)/R
where, for example, P is expressed in watts, I in amps, R in ohms, and E in volts.

Therefore E = (Square root (P x R)).

So for example 100 watts into an 8 ohm resistive load corresponds to:

Square root (100 x 8) = 28.28 volts.

Regards,
-- Al

Yes, RMS = root mean square. As you probably realize, the signal provided to a speaker consists of various frequency components each of which is AC (alternating current). Amplifier power capability is defined based on the simplified assumption that the signal consists of a pure sine wave at a single frequency, with that single frequency being anywhere within some range of frequencies, such as 20 Hz to 20 KHz. The RMS value of a sine wave equals its peak (maximum) instantaneous value divided by the square root of 2, or approximately 0.707 x the peak value.

In audio voltages and currents are usually defined on an RMS basis, in part because the amount of power supplied to a resistive load that can be calculated based on RMS voltage and current numbers (even for waveforms that are not sine waves) equals the amount of power that would result from a DC (direct current) voltage and current having the same values, that amount of power in turn being proportional to the amount of heat that is produced when supplied to ("dissipated in") a resistive load.

By the way, one thing that often causes confusion in this context is that the word "peak" can be used to mean two different things. It can refer to the peak (maximum) value of a sine wave or other signal at any instant of time during each of its cycles (corresponding for a sine wave to the RMS value divided by 0.707), or it can refer to the peak (maximum) RMS value that can be reached by that sine wave or other signal during normal (or other) operating conditions.

Kudos for your interest in these matters. Regards,
-- Al

Jim, yes, any frequency other than zero Hertz (which is DC) is alternating current. And of course a music signal nearly always consists of a mix of a great many AC frequencies that are simultaneously present, at a wide variety of "amplitudes" (i.e., strengths, or magnitudes).

why can’t the resistance from the speaker, that the amp relies on be built into the amp itself?

In order for a speaker to absorb electrical power, some fraction of which it converts into sound, it has to have resistance. And for it to absorb a reasonable amount of power when provided with voltages that are reasonably practical, that resistance has to be relatively low (e.g., in the vicinity of 4 or 8 or 16 ohms or so). If a similar resistance were placed into the amp, the resistor in the amp would absorb power but convert it into heat rather than sound. So that resistor would serve no useful purpose, but would reduce the amount of power the amp would be capable of providing to the speaker.

Best regards,
-- Al

Hi Jim,

As a consequence of the equations that were cited earlier for the relations between power, voltage, current, and resistance, it can be inferred that delivering a given amount of power into 8 ohms requires much more voltage and much less current than delivering the same amount of power into a very low impedance, such as 1 or 2 ohms or less. While conversely delivering a given amount of power into a very low impedance requires much more current and much less voltage than delivering the same amount of power into 8 ohms.

And, hypothetically speaking, if the load impedance were to truly approach zero (i.e., a true short circuit), the amount of current required to provide any voltage and deliver any power would approach infinity.

For an amplifier to be able to deliver amounts of power that are generally desirable into both 8 ohms and very low impedances it therefore has to be able to supply BOTH very large amounts of current and relatively high voltages. To be able to do that it will have to be much larger, heavier, and more costly than would otherwise be the case. And most likely sonic compromises would result as well.

There are some amps that can supply substantial amounts of power into impedances of 1 ohm or thereabouts, and in a few cases perhaps into even lower impedances, but in all of those cases I am familiar with the amps are big heavy monsters, which consume large amounts of electricity, generate a lot of heat, and don’t necessarily sound as good as many other amps that are in the same price range.

Regards,
-- Al