Let me start from the easiest of your questions: power consumption is largely irrelevant to the issue (except... more on that later).
The relevant value for E (or V) is the internal voltage of the amplifier power supply, not what comes out of the wall socket/receptacle. That 'E' is specified to provide a certain amount of power to a load, which in domestic audio is generally taken to be an 8Ω resistor. Hence the typical specification of (say) 50 Wpc/8Ω.
Note that in a resistor, electromotive force and current are linked by Ohm's law, which says:
E = I * R (where R is the resistance)
now, if you replace that into the 'PIE' equation, you get:
P = E²/R = I²R
so, assuming a load of 8Ω, a 50 wpc amp has to provide
E = √(50*8) = 20 V of internal voltage
I = √(50/8) = 2.5 A of load current
So far, so good; the designer specifies a power supply and final stage that can provide 2.5 A and 20 V, and everyone is happy.
The problem - and it can be a problem - is that loudspeakers are not resistors: the 'resistance' (more appropriately impedance) they present to an electrical current varies with the frequency of the input signal, and it can be much lower or higher than the nominal (most commonly 8Ω or 4Ω) at certain frequencies.
For example, let's assume that speaker X has a minimum impedance of 2Ω at 200 Hz, vs. a nominal impedance of 8Ω (usually at 1 kHz). What happens to the current then, when the amplifier is fed a 200 Hz input signal such that the output is driven to its maximum?
Well, the voltage is still going to be 20V, and Ohm's law still applies:
I = E/R = 20/2 = 10A
That's four times our initial specification. This can spell trouble for both the power supply and the final stage, particularly if this is sustained over significant periods of time. The designers can take precautions (protection circuitry) to prevent damages, but that may mean sound quality degradation and/or shut-down.
Alternatively, the designers can incorporate in the specification the ability to provide (at least temporarily) much higher currents than the standard, continuous Wpc/8Ω would imply. This is a 'high current amp'.
Where does 'consumption' power come in? Well, typically that is calculated as the maximum (continuous) power that the amp will draw - note that in the case above (E = 20V, R = 2Ω), P = E²R = 200W
If the amp is capable of providing indefinitely 200W to a 2Ω load, then the consumption power has to be at least that much (plus inefficiencies and 'overhead' power needed to manage other functionality e.g. power meters, lighting).
Finally, an example of a high current amp - Accuphase E800 (50 Wpc/8Ω; 100 Wpc/4Ω; 200 Wpc/2Ω; 300 Wpc/1Ω). A low current amp - AudioNote P1 (9 Wpc/8Ω; 9Wpc/4Ω; no information on lower loads, but, given the design, power is likely to go down rather than up). Note that what matters is not the 'starting number' of Wpc, but how much they go up - or not - with decreasing load impedance.
There's a lot more to explore on amps design and power specifications, and the above is necessarily approximate, but I hope this short novel at least answers your initial questions.