Why doesn't 1 Watt = 1 Watt?

Factual Questions
1

I work with high powered speakers, and these systems use car batteries to generate the amps needed to crank out the watts. What I cannot understand is why the equation W = VI isn’t good enough to make these speakers work! Humpf!

For example, 100W could be produced by 120VAC, but then only 0.83 amps would be needed. Yet, using two car batteries series, I produce 100W with 24V using 5 times the amps??? Physics never taught me anything more than W=VI!!! How can one determine how many amps one needs??? Any number of combinations of V and I will yield the necessary wattage!!!

So, when does 1 watt <> 1 watt? Why should electricity care HOW I got the 100W, as long as I can produce 100watts, right? Apparently not… :mad: But, why??? This is illogical, Captain! :dubious: Grrrr…!

2

Oh, electricity cares. More than you will ever know.

3

You also need to know Ohms law, V=IR, where R is the resistance. The higher the resistance, the more voltage you need to get the same current. With Ohm’s law, we can compute the power P and get P = VI = I^2*R or = V^2/R. This would be the case if the speakers presented a fixed resistance R, like 8 Ohms, in which case the power is Voltage squared over 8 Ohms, or current squared times 8 Ohms. Unfortunately, life is more complicated than this, since the resistance R is actually dependent on frequency and in general is a compex number (we call it impedance when we have to take account of the imaginary component of the resistance). Aren’t you glad you asked?

4

Your math looks good, both situations yield approximately 100W. The equation is correct, but that is pretty much it. Electronic components can be pretty picky on what they accept, though, and wattage most likely isn’t the only requirement on the input.

Have you considered that 120VAC is, of course, AC, while car batteries yield DC?

There’s nothing in physics that says that two phenonmena that have the same measurement on some scale are always identical. Consider a lamp that gives out a certain lumen rating of general fuzzy white light. Then, consider a laser that yields the same luminosity rating but releases a thin coherent beam of a specific frequency of red light.

5

20 V DC can be converted to 40VAC.

6

If you are measuring 24 volts and 5 amps at the source, and 100 watts of power at the load, then it sounds to me like 20 watts is being spent heating up the amplifier. There is a fudge factor here since I assume you are measuring RMS power at the load and we’re starting off with steady state DC power at the source. That’s not a straightforward calculation that you can do on the back of a bar napkin.

And since you mention 120V AC, I assume you’re using a power inverter and run a household amplifier which, in turn, has its own internal power supply that converts everything right back to DC again? There’s a whole lotta waste going on there, too.

7

I’m guessing the main issue here is impedance. Setting aside the more complicated frequency-dependency and inefficiency issues for the moment, you can picture a graph of how your speakers will accept power. Draw current in amps on one axis and potential in volts on the other. What your speakers can do will appear on a straight line going through the origin, and the slope of the line is your speaker impedance. What the entire universe of possible hypothetical 100 W power supplies can do will be a curve that asymptotically approaches the two axes but is running from southeast to northwest at the point (10,10). This curve is a hyperbola. Where the speaker line and the possible power supply curve intersect, you will be pumping 100 W through the speakers. However, if you use some different power supply from somewhere else on this curve, it will be unable to deliver 100 W to the speaker because their impedances are poorly matched.

One example of an impedance mismatch would be a motorcycle and a farm tractor that both have 50 horsepower engines. The motorcycle will just stall if you try to pull a plow, and the tractor will never make it up to the minimum allowable speed on that entrance ramp to the highway. They each can do the work of 50 horses, but in different speed regimes.

8

I read the OP five times. I apologize, but I still don’t understand the question.

P = VI. This equation is not an approximation; it is absolute. More specifically, we can talk about instantaneous power, average power, and peak power. The definitions of each are self explanatory.

If two car batteries in series are required to transfer energy at the rate of 100 watts *continuous *(“DC power”), then the current will be 4.17 ADC. If two car batteries in series are required to transfer an *average *power of 100 watts, then the current will be 4.17 A[sub]RMS[/sub]. If two car batteries in series are required to transfer a *peak *power of 100 watts, then the peak current will be 4.17 A.

The instantaneous amount of power going to a speaker is the instantaneous voltage multiplied by the instantaneous current. The maximum instantaneous power going to a speaker depends on a lot of variables, the most significant simply being the design of the amplifier. It should also be mentioned that it is possible for the instantaneous amount of power going to a speaker to be higher than the average power being transferred from the batteries. This is due to energy storage devices (caps and inductors) inside the amp.

The average power going to a speaker is the RMS voltage across the speaker multiplied by the RMS current through the speaker. (Sometimes this is called “RMS power”, though I hate that term because it implies we are mathematically calculating the RMS of the power waveform. In reality, we don’t care about the RMS of the power waveform, only the *average *power.) The average power going to a speaker will be less than the average power being transferred from the batteries due to inefficiencies in the amplifier. (The amplifier generates some heat, after all.) In addition, the magnitudes of the peak voltage and average voltage going to the speaker have little to do with the battery voltage. The amplifier likely has a SMPS that boosts the 24 VDC battery voltage to a something around 100 V or so.

9

OP appears to be under the impression that superconductors and 100% efficient power amps are easily and cheaply acquired and mainstays in the field of consumer electronics. :smiley:

10

And before I get “called” on it, I guess I should state that the average power is also a function of the phase angle between the voltage and current.

11

Good save - most of us were thinking how to phrase our witty retorts to this gaffe - you could almost hear the collective sigh of “Awww” when you closed the window on us.

12

LOL!

Well, you know how these techie threads get. Gotta CYA. :smiley:

13

If I understand you correctly, you’re asking why you can’t supply a 100 watt device with any combination of VI=100. The short answer is that you don’t supply a set voltage and current - generally you supply a set voltage and the current is a consequence of that voltage and the device’s impedance. If you try to supply a higher current, you have to supply a higher voltage, not a lower one, and so the power you supply is increased.

If you ride a bicycle, think about the gears. You can go the same speed by pedalling really fast (without much pressure) in a low gear or pedalling really slow (but pushing hard) in a high gear. The power is the same, but you can’t just decide to push hard and go slow in the low gear - it doesn’t work. The pedalling rate is a consequence of how hard you push AND the gear ratio. Similarly with electricity, the current is a consequence of the applied voltage and the impedance.

14

I too am kind of confused about what the OP is asking, but I thought I’d muddy the waters just a bit more by talking about apparent power and reactive power and all that.

See, the equation P=VI only works on instantaneous values of V and I, and it only provides the instantaneous power. Often when discussing sinusoidal AC systems, we are more interested in average power delivered to a load, in which case there are 3 different kinds.

There is apparent power, S = VI*, where V and I in this case are complex quantities called phasors which have both a magnitude (the RMS value in the case of V and I) and a phase, and the star denotes the complex conjugate. {See, every sine wave has a magnitude and a phase shift (measured in degrees or radians) from some reference phase. We usually assign the voltage at one point in the circuit a phase of 0, and measure the phases of the other voltages and currents relative to that. Complex cojugation in this case is simply a matter of negating the phase (+30 degrees becomes -30, for instance).} Apparent power is measured in Volt-Amperes, or VA.

There is real power that we are used to, P = VIcos(theta) where theta is the phase difference between V and I. The cos(theta) part goes by the name of “power factor”, and takes on values between 0 and 1. Since the power company can only charge you for real power, they like the power factor to be as close to 1 as possible. Real power is measured in Watts, W.

Finally, there is reactive power, Q = VIsin(theta), measured in Volt-Amps-Reactive, or Var. This is the part of the apparent power that does no work. It is produced as a consequence of the inductance or capacitance of the load. If Q is positive, it is inductive, if negative, it is capacitive. Capacitive (negative) Q flows out of the load rather than into it.

Apparent power, also called complex power, is the sum of the real and reactive power, S = P +jQ, where j is the imaginary unit.

Anyway, I may have written too much, but the point is that when we look at average power in AC systems, things become a lot more complicated than p=vi. The complications arise because of the reactive impedance in the circuit. If the circuit is purely resistive, then there are no phase shifts, Q becomes zero, and P = S = VI. This is another reason why the power company likes a power factor of 1.

15

What DrCube said just now. One also has to apply a bit of intelligence as to what exactly is happening to this energy flow - is it being dissipated as heat, light, sound, radio waves etc? Digression: Take the case of an AC signal flowing through a perfect capacitor, plot instantaneous power against time, and then work out why the capacitor doesn’t get hot like the graph suggests.

With the case of speakers and amplifiers, the power ratings are invariably quoted in Bullshit Watts. Reputable (i.e. very few) manufacturers will quote the power figure as a (relatively) undistorted sinewave at 1 kHz frequency into a specified speaker impedance. But most will use whatever underhand technique gives the biggest figure. For instance, a loudspeaker won’t have a flat impedance characteristic wrt frequency, it will look like a mountain range, and the lowest impedance point will invariably be at the speakers self-resonant frequency. Use this frequency for power tests and the figure starts looking better already. Not being too concerned about clipping the waveform squeezes out a few more BS watts from the marketing cloaca, and if you’re only measuring the few cycles at this load before the reservoir caps start to sag then even better.

The OP makes a good point about the voltage - there’s no way around it, for a fixed loudspeaker impedance you need to deliver the volts to get the power, and a 12V car battery won’t deliver much power into a standard 4 ohm car speaker. It’s possible to make the speaker lower impedance, but then a lot of power would be lost in the finite-resistance wiring and imperfect connectors, so a standard approach is to raise the voltage headroom of the power amplifier by a voltage-boosting power supply stage. This is usually integral to the amp unit.

A simpler way of doubling the voltage to the speaker (and hence quadrupling the power) is to connect the speaker as a bridge-tied load. Here the negative terminal of the speaker is connected to an amp that produces the exact inverse of the signal sent to the positive terminal of the speaker. This is a very popular technique with car stereos, but not really used with domestic hi-fi amps as the home AC supply enables the use of transformers to set the available voltage.