Why are low impedance speakers harder to drive than high impedance speakers

Al, THANK YOU very much! I truly appreciate your full, detailed response that also went to the heart of the issue. I think I am beginning to understand. It seems like the critical part (at least for me) is the following:

  Put simply, it is easy for an amp to supply voltage, as long as it is operated within the range of voltage it is capable of, but less easy for it to supply current.

Ohm’s law – by itself - doesn’t seem to get at this.

If I truly understand how the relationships work, there are several steps involved:

1. Volts (voltage) x Amperes (current) = Watts (power)
2. Ohms Law: Amperes (current) = Volts (voltage) / Ohms (resistance or impedance)
3. The amount of Watts or power required to drive a speaker of a certain efficiency to a certain SPL in a certain space remains constant even if you change a speaker’s impedance.

The following example demonstrates the relationship:

• A) 2 Amperes = 16 Volts / 8 Ohms where 2 Amps x 16 Volts = 32 Watts
• B) 4 Amperes = 8 Volts / 2 Ohms where Amps x 8 Volts = 32 Watts

Therefore, when you reduce impedance, but keep power constant, the current increases but the voltage decreases. This is where the crucial piece of information applies to clarify that the reduction in voltage does not mitigate the increased energy required by an amp to increase the current.

Am I close????

PS.  Still not sure how the pipe analogy works here?

Am I close????

You’re better than close; that’s exactly right :-)

I’ll mention also that the following equations can be derived by substituting some of the terms in equation 2 in your post above into equation 1, and doing some algebraic rearrangements, and these equations may add some further clarity to what has been said:

Power (watts) = (Volts squared) / Ohms

Power (watts) = (Amperes squared) x Ohms

It can be seen from these equations that for power to remain constant, as the number of ohms decreases voltage must decrease, while current must increase.

Finally, to be precise I should mention that we’re simplifying all of this a bit by making the assumption that the load is purely resistive. Volts x Amps = Watts in the case of a resistive load, but things get somewhat more complex when the load has a significant inductive or capacitive component, in addition to its resistive component.

Regards,
-- Al

To me it seems like if you need twice the amount of electrons to flow to a 4 ohm speaker than to an 8 ohm speaker, the amplifiers would need to work harder and in return wouldn't this cause more distortion?  Also, if it takes twice the amount of current to drive a 4 ohm vs an 8 ohm speaker, wouldn't this mean the 4 ohm speaker is less efficient.  However, you would think the 8 ohm speaker would be less efficient because it has twice the amount of resistance to the current.

I was told once a 4 ohm speaker requires twice the amount of current than an 8 ohm speaker.  Because a 4 ohm amplifier delivers twice the amount of current, would this in turn supply twice the amount of information to the speaker to create more detail in the music?

Since the OP thinks in mechanical terms, this analogy may best describe how speaker loads affect an amp.
Imagine the amp is a somewhat fragile flywheel that will fly apart at a certain RPM. If you apply a resistance (an 8 ohm speaker load in electrical terms) to the flywheel that is sufficient to prevent the flywheel from reaching critical speed it will not break. If you remove part of the resistance (think 4 ohm speaker) the flywheel will speed up. If you remove enough of the load, you eventually reach a point where the flywheel is spinning so fast it fails. Replace the concept of the flywheel speed with power output from the amp. The power an amp will produce is inversely proportional to the resistance to current flow. More resistance keeps the amp in check so to speak.