I'm sure everyone knows the formula; Watts=Volts*Amps

Does this apply to both AC and DC?

 

No. Conventionally (at least where I am), I = current, P = real power

DC:
P = VI

AC:
Vrms = Vpeak/sqrt(2)
Irms = Ipeak/sqrt(2)
P = Vrms*Irms = [Vpeak/sqrt(2)]*[Ipeak/sqrt(2)] = 0.5*Vpeak*Ipeak

 

Ok, so what do the variables for the AC represent? Will you show me an example using this information;
Voltage = 120vac
Amprage = .8amps

I need to know how many watts it takes.

 

If the 120V is the peak voltage of the sinusoid AC signal and 0.8A is the peak current, then you go:

P = [120/sqrt(2)] * [0.8/sqrt(2)]

or you can combine the sqrt(2) into one number and just do:
P = 0.5*120V*0.8A

If 120V and 0.8A are in RMS, then you just go
P = 120V*0.8A

It's pretty much all the same thing, just the root(2) is sometimes combined into an RMS number or the two root(2) that are multipled are combined to be written as 2.

 

Power in n AC circuit is measured in Volt Amps because the only real power is in the resistive element. This has always been a sore point with me in the USA because they describe AC generators in terms of Watts which of course is nonsense. The apparent power is what you measure when you multply current by volts and current depends on the none resistive elements in the circuit. That is why it is important to add capacitance to an AC circuit because most loads are inductive.

 

So then the laptop(that is what the numbers were from) should take about 48 watts of power? Thanks!

 

Yes. Assuming 100% efficiency for the components you did not take into account (like AC-DC converters).

 

In AC, you have the "real power (W -> Watts)", that is the power developed in a purely ohmic-resistance circuit.
And you have the "reactive power (VAr -> Volt-Ampère reactive)", that is the power developed in a inductive/capacitive-reactance circuit.

The W and VAr forms the "Apparent Power (Volt-Ampère)."

They form a rectangled triangle, where the Real Power and Reactive Power form the legs, and the Apparent Power forms the hypotenuse.

P.GIF

 

So AC can be graphed in linear form?

 

one question that will simplify your answers (AC power is fun!!!)

what kind of current is being drawn?

if it is sinusoidal and in phase with the voltage (ie just a resistive load) then take the RMS voltage and RMS current and multiply them together

IF the load is inductive/capacitive then things start getting complicated (and VA start becoming relevant)

equally if the load is non-linear (say a diode-bridge rectifier) it again gets complicated

__________________
Nothing is impossible.
Once a problem is realised, the rest is just details

 

watts=VI cos(angle)...where it is the phase angle. the AC cant be plotted as a liner wave.its only the rms value not the complete waveform.

 

now is getting complicated. I dont know what inductive,or the other things mean. What do they mean?

 

It has to do with components in the circuit storing energy in different ways (capacitance and inductance) which causes the current AC sinusoid waveform and the voltage AC waveform to be out of phase (the sinusoids don't look the same), which affects the apparent power which is dependent on the current and voltage at that instant. Shifting the waveforms causes the voltage and current to be at different values in the same instant than if both waveforms were in phase.

It's kind of something you have to read a book for. You need to know about complex numbers.

 

So there are very small spikes?

 

They're not spikes. Through a purely resistive load, the current is at zero/max/min whenever the voltage is at zero/max/min. But inductive loads make it so that the current lags the voltage (the current reaches zero/max/min after the voltage does).

And you can see that this means that the power isn't the same anymore. For example, in a resistive load when current and voltage peak at the same time, that's when there is the most power since V(t=T)*I(t=T) is largest since max V and max I occur at the same time and are multiplied together, but if the current or voltage is shifted a little bit, the V*I is smaller, since now the when the maximum of the current and voltage no longer occur at the same time, so the highest V*I that you can achieve will be lower.

Imagine two identical sychronized sine waves graphed out, and multiply them together to make a 3rd graph. Now imagine that you shifted one of the first two sine waves a bit. Now multiple the shited and unshifted graph together. The graph made from two synchronized sine waves (kind of representative of power) will have higher peaks and different values than the graph made from unsynched waves.