There appears to be a typo between your first post and your last, but no matter. I think the point made in this thread is that there are too many variables that prevent giving a pinpoint explanation for the different current levels.
An amp is an amp, but VDC is not the same as VAC. Nor is resistance the same as reactance. So they are not all interchangeable. Just ask Chris. I think the simplest solution, again, is to multiply the power needs of one bulb by the number of bulbs you're going to use.
If the supplied transformer is 60w and the dc one is only 18w you may be pulling the voltage down so that the current is less. Check the voltage and current of both supplies at load.
The bulb rating of 30 watts at 12 volts is most likely the manufacturer's AC-derived spec. The current formula for that is power divided by (voltage times the cosine theta). The DC Ohm's law I=P/E doesn't work in this case.
One amp is one amp. <p>How you count one amp is where you get varying numbers and ideas. RMS, etc.<p>With AC you can only count the amps, under the curve. <p>Every thing out side the curve is ZERO. <p>And then you have time to add into the factor, and is it a Sine wave, triangle, square, etc, when dealing with AC.<p> The actual amp is comprised of X amounts of electrons, something like 10 to the 8th power of electrons is one amp? [approximately 6.241 506 x 1018 e]<p> When you have a changing wave form, whether they are pushed and then pulled through the gate, or just on and off for time periods, the increase and decrease doesn’t allow a straight sum to be counted because of the ramping or off time over the cycle. <p>So the amperages are a variable over time, with points in the wave form where no current is flowing and times when maximum amperages are flowing. <p>That’s why we have elaborate algebraic schemes to count the total variable over time and turn it into a simple term, and treat it as if it were DC. <p>Whether the electron moves forward, and then moves back, or just moves in one direction, work is still done and this accounts for the AMPs used.<p>[ March 14, 2005: Message edited by: Chris Smith ]</p>
Kill another variable: use a 12 V car battery for your DC measurements. Wall warts are notoriously sloppy in waveform, regulation, and just plain output voltage.<p>A good "stiff" (low impedance) DC current source like a car battery will eliminate the possibility that your meter is being fooled by the sloppy waveform of the wall wart.<p>I've got $10 that says this is the main source of your measurement anomolies.<p>As I said,<p>"..you are measuring Average Values when you measure the DC current, but very close to RMS values when you are measuring the AC current, which should be close to a pure sine wave."<p>I would modify that as follows to emphasize the wall wart variable:<p>"...you are measuring Average Values when you measure the DC current of a sloppy, chopped up wave from the supposedly pure DC output of the wall wart, but very close to RMS values when you are measuring the AC current, which should be close to a pure sine wave."<p>Make that $50.<p>Any takers?<p>[ March 15, 2005: Message edited by: terri ]</p>
The lamp setup like you descibe, many times use switching supplies that actually out put hi-freqency AC (5-10 Khz).This would give screwy readings in either AC or DC on your lo-frquency meter.
Delete one of your posts. That happens to me a lot if I use the back button to get back to the main board, but instead it sends me back to the "Post A Reply" screen and then I hit "Add Reply" again.<p>Click "Edit" (the icon with the pencil over each posting) and the "Delete Post? (check box, if yes)" will show up in the upper left corner of the edit screen. <p>Check this box and hit "Edit post" again. This will remove the duplicate posting for you. <p>Don't post a thank you, because after you delete the duplicate posting, I'll erase this post and then your thank you response will look goofy just hanging out there by itself until you erase it, too.<p>[ April 01, 2005: Message edited by: terri ]</p>
I think the simplest solution, again, is to multiply the power needs of one bulb by the number of bulbs you're going to use.
<p>Don't post a thank you, because after you delete the duplicate posting, I'll erase this post and then your thank you response will look goofy just hanging out there by itself until you erase it, too.<p>[ April 01, 2005: Message edited by: terri ]</p>