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

Would it be possible to go back to the original question? This thread contains lots of great information, but I am still struggling with the basics. I apologize for this “newbie” request, but I'm clearly misunderstanding something and would be grateful for some help!

All other things being equal (PLEASE SEE full caveats note at bottom), I thought the following were generally true:

1. If you double an amp’s wattage, you increase your potential spl by 3db. Thus, an amp which doubles its watts when the speaker load is halved (not all amps truly do this) driving an 8 ohm speaker with a sensitivity of 90db would generally have essentially the same spl potential driving a 4 ohm speaker with a sensitivity of 87db.
2. In general, the higher you turn up the volume of your amp to the same set of speakers, the harder it has to work. All the way up is relatively hard work (and may cause clipping, etc.), barely on should not be much work.
3. Theoretically, if the same speaker system could be made in an 8 ohm version and a 4 ohm version with the same sensitivity rating (let’s say 90db), then the same amp driving the 4 ohm version would need less “signal” or a lower volume setting to attain the same spl in the same room as the 8 ohm version.

If the above are basically right, I don’t understand why an amp would need to work harder with a 4 ohm load than an 8 ohm load to put out the same spl in the same room.  If the above are not correct, where did I go wrong?

THE HOSE ANALOGY: I’ve heard the previously referenced analogy of water going through a hose many times, and each time it sounds backwards to me. It seems to me that if the flow of water – or amount of water moving through space over time – is to remain constant, a larger diameter pipe, or lower impedance, would make it easier for a pump (or amp) to push that water.  If the flow remains constant, then as the hose diameter decreases, the pressure increases and the pump would need to work harder. Why is this not correct?

Clearly, I must misunderstand some fundamental concepts! I’m not an engineer or science “type” so may need some baby steps.

PLEASE NOTE: in order to try to understand the basics, all of the above is based on simplistic and theoretical situations, with all other things such as speaker configuration and design, other components, room size, etc. being equal AND with all components properly matched.  I understand that real world implementations may vary.

Hi,
If you keep the Ohm’s law equation in mind it’s easy to explain. I = (current), V=voltage and R = resistance (speaker load impedance expressed as ohms).
I=V÷R. So the smaller R becomes, the larger I becomes.
If for example V=10 and R=10 then I will =1. If R is reduced to 1 then I now =10. Reducing the R (speaker impedance expressed in ohms) will increase the value of I (current demand). So a 2 ohms load (lower R) will demand more current  (I) than a 8 ohm load  (larger R). 
Charles

Thanks for the reply.  Could you please relate this to my 3 points above and the hose analogy as well?  I see the formula, but don't know how that applies to needing less signal for an amp that can produce more watts because there is less resistance.  How does that relate to a pump needing to work harder (or less hard in the amp world) to push the same amount of water through a smaller diameter hose?

Sorry, I thought Ohm's  Law is self evident in its equation.  As R decreases I must increase for a given V  I.e. voltage. There's an inverse relationship,  more current is needed to maintain the voltage as the resistance diminishes. It all seems straightforward to me.
Charles 

I’ll add a few comments to Charles’ excellent answers.

Swingfingers, first let’s change the word "sensitivity" in your question to "efficiency." Speaker sensitivity is usually defined on the basis of an input to the speaker of 2.83 volts, rather than 1 watt. 2.83 volts into 8 ohms corresponds to 1 watt, so the resultant SPL (sound pressure level, in db) is the same either way. But 2.83 volts into 4 ohms corresponds to 2 watts, so if the 87 db figure you referred to for the 4 ohm speaker is defined on the basis of a 2.83 volt input that speaker would produce only 84 db in response to 1 watt.

So with the word "sensitivity" (which we’ll define as db SPL at 1 meter in response to a 2.83 volt input) changed to "efficiency" (which we’ll define as db SPL at 1 meter in response to a 1 watt input, although in some other contexts the term "efficiency" may also be used to refer to the ratio of acoustic power out to electrical power in), my answers to your three questions are:

Q1)Yes.

Q2)Yes, with the slight qualification that in the specific case of a class A amplifier the amp will dissipate (consume) less power internally (and therefore have a lower internal operating temperature) when it is supplying large amounts of power to the speaker than when it is supplying small amounts of power (or no power) to the speaker. And in that sense and to that extent (there are other factors that come into play, of course) a class A amp may be working less hard when supplying more power rather than less.

Q3)Yes, a 90 db/1 watt/1 meter/4 ohm speaker will require a lower setting of the volume control to produce the same volume as a 90 db/1 watt/1 meter/8 ohm speaker.

If the above are basically right, I don’t understand why an amp would need to work harder with a 4 ohm load than an 8 ohm load to put out the same spl in the same room. If the above are not correct, where did I go wrong?

Keep in mind that the speakers referred to in Q2 are identical, while in Q3 they are not.

In both situations referred to in Q3, the amp will deliver the same amount of power to produce a given SPL. For a resistive load power = voltage x current. The volume control setting controls the amp’s output voltage, while the impedance of the speaker determines how much current is drawn from the amp at a given output voltage. In the case of the 4 ohm speaker the lowered setting of the volume control that you correctly referred to will result in less voltage being supplied by the amp compared to the 8 ohm case, but the amp will be supplying more current at that lowered volume control setting than at the higher volume control setting of the 8 ohm case. 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.

I’ll leave the hose analogy question to others, as I generally prefer to avoid using non-electrical analogies for electrical things.

Hope that helps. Regards,
-- Al