| itsme.ultimate@gmail.com 2007-07-13, 5:25 pm |
| X-No-Archive:Yes
Battery and resistor. You measure the voltage across the battery and
multiply it by the voltage across the shunt and the constant and you
have power.
I'm confused about how I should interpret the data of the real life
measurements I've taken. Power source is two 12v nominal batteries in
series, which gives around 25v under load and 0.5v ripple.
Load: high frequency modulated sinewave UPS which generates sinewave
using high frequency modulation then sends the output to step-up
transformer to yield 120v rms.
The input current waveform is very variable and looks like this:
http://img58.imageshack.us/img58/14...entwavepf8.jpg. When I
expand the time base, I can see the high frequency modulation
The voltage at battery was ~25v with 0.5v ripple under load, so I'm
going to call that clean DC. DC ammeter reads about 3.2A, which means
around 80W using simple math.
50KHz bandwidth power analyzer(should be enough to not overlook the
high frequency component from sinewave synthesizer) reports 80.3W,
which makes the simple math look right, but it reports 5.333A
133.32VA.
The analyzer takes the square root of the sum of the squares to get
the current, so
sqrt(3.2ADC^2+4.26625AAC^2)=5.333A
Real question is, how much current through that shunt? 5.333A or
3.2A?
What is the effect on the available capacity of battery? Do I
calculate as if the load is 3.2A or 5.333A?
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