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An everyday analog (D'Arsonval) movement automatically averages a waveform.
Suppose the triangle peaks are 1 A, troughs at zero. The average is 1/2 A. This is what your meter will read. (It can be derived geometrically based on the formula for the area of a triangle.)
When measuring current, the meter inserts a certain amount of resistance. A different resistor for each range. The highest ranges insert lesser ohms.
You must select the highest ampere range which gives a usable reading
Thanks for your response. This applies to extreme waveforms too then correct? Such as a 1200V sawtooth like the one measured here on my scope? This is being discharged from a .33uF capacitor, I'm trying to get a general measurement of the current.
I don't seem to have a way to adjust anything on my cheap DC analog ammeter.
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