effective power

[dudewheresmymd]

anyone know where they got this formula?

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[DrknoSDN]

The average value of a sine wave is going to be 50% it's peak value. Also:
Watts = Amps * Volts
and V = IR
So... Watt = I * (IR)

3200/2 = I*I*1
or sqrt (1600) = I

 

[chenzt]

The average value of a sine wave is going to be 50% it's peak value. Also:
Watts = Amps * Volts
and V = IR
So... Watt = I * (IR)

3200/2 = I*I*1
or sqrt (1600) = I

Actually, RMS average of a sine wave is ~0.707 or root(0.5). Since we're talking about current here, when calculating power, we need to square it.

W = I*I*R

Average I = 0.707 of maximum

W = 0.707 x 0.707 x 1 = 0.5

 

[DrknoSDN]

What thinking of a sinusoidal power graph. As current and voltage go negative they still produce positive power. Using RMS current much more reliable for mcat tho.

Pictures--> http://www.ee.sc.edu/personal/faculty/simin/ELCT221/22 AC power.pdf

 

[DrknoSDN]

Actually, RMS average of a sine wave is ~0.707 or root(0.5). Since we're talking about current here, when calculating power, we need to square it.

W = I*I*R

Average I = 0.707 of maximum

W = 0.707 x 0.707 x 1 = 0.5

chenzt, after a lot of thinking about your post (mostly thinking that I was stupid for not using RMS to derive the equation), I don't believe your using those equations correctly. I kept trying to understand your method but I cannot find a way to make RMS useful here.
You say W = I*I*R and W = 0.707 x 0.707 x 1 = 0.5
That is true, but only because the problem provided a resistance of 1 ohm.

With Max power of 3200 W and resistance of say 25 ohm.
0.707 * 0.707 * 25 ohms = 12.5 ?
(not very useful because the Average power is not 12.5x the max power)

I think the better way to solve would be to just know that Average Power is 1/2 max or (1600), divide by resistance of 25 and solve for I.
(3200/2) / 25 = I^2
I = sqrt(64) = 8 amps

I believe that's right because AC current changes inversely to the square of the change in resistance. So a 25 fold increase in resistance is a 5 fold decrease in AC current. 40 ->8
If anyone can let me know if they have a better way to solve, or if I'm totally incorrect (very possible) I would appreciate it.

 

[SpaceBuff12]

You say P = I*I*R and Pavg = 0.707 Imax * 0.707Imax * 1 = 0.5*(Imax)^2
That is true, but only because the problem provided a resistance of 1 ohm.

With Max power of 3200 W and resistance of say 25 ohm.
Pavg = 0.707 Imax* 0.707 Imax * 25 ohms = 12.5*(Imax)^2?
(not very useful because the Average power is not 12.5x the max power)

If you increase your resistance and the max power stays the same then the max current decreases.

 

[chenzt]

chenzt, after a lot of thinking about your post (mostly thinking that I was stupid for not using RMS to derive the equation), I don't believe your using those equations correctly. I kept trying to understand your method but I cannot find a way to make RMS useful here.
You say W = I*I*R and W = 0.707 x 0.707 x 1 = 0.5
That is true, but only because the problem provided a resistance of 1 ohm.

With Max power of 3200 W and resistance of say 25 ohm.
0.707 * 0.707 * 25 ohms = 12.5 ?
(not very useful because the Average power is not 12.5x the max power)

I think the better way to solve would be to just know that Average Power is 1/2 max or (1600), divide by resistance of 25 and solve for I.
(3200/2) / 25 = I^2
I = sqrt(64) = 8 amps

I believe that's right because AC current changes inversely to the square of the change in resistance. So a 25 fold increase in resistance is a 5 fold decrease in AC current. 40 ->8
If anyone can let me know if they have a better way to solve, or if I'm totally incorrect (very possible) I would appreciate it.

What I meant was 70.7% or 0.707 times the maximum current. This is because average current of an alternating current is RMS of maximum current.

W = I*I*R, R is irrelevant, since it is the same resistance whether current is at max or average.

At max current (we'll call it 1 because it is easy to calculate, but you can choose any current you like, such as 2, 3, 4, etc) and a resistance of 1

W = 1*1*1 or 1 Watt (or if we use Imax = 2, W = 4)

At average current, 0.707 * 0.707 * 1 = 0.5 watt, or half of max power (use Imax = 2, then it is 1.414 x 1.414 x 1 or 2 watt, still half of max power)

That's why the solution manual made the statement that avg power = 1/2 of max power, which I think what the OP was asking. Your post was pretty accurate as to how to solve, I was just adding to the why.

I guess I could've made it a little bit more clear