I'm trying to find how the internal resistance of a battery changes over time (i.e. see how it changes as it goes flat.)
Now I know what I 'could' do to find out how the internal resistance changes over time, I could run the battery and at given intervals change the resistance several times quickly (recording both the voltage and the current) then returning to the original Resistance. This allows me to draw a 'Voltage Vs Current' graph for each time interval and using the gradient of that graph as the internal resistance value. However, this method would be unbelievably time consuming - and as the resistance in the circuit wouldn't always be constant (although it almost would be as I would change it very quickly) the results wouldn't be accurate.
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File Format: PDF/Adobe Acrobat - View as HTMLcurrent needed to indicate FSD, and its internal resistance. Now using Ohm’s law , it’s easy to work out the voltage. drop across the meter when it’s
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File Format: PDF/Adobe Acrobat - View as HTMLCalculate the current through each resistor using Ohm’s Law and .. voltage of the LVPS is above 12.0V , the internal resistance increases substantially.
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Would it be possible to work out the internal Resistance of a battery purely by having a graph of Current over time and a graph of Voltage over time.
Resistor= 1ohm
Current - Goes down over time
Voltage - Goes down over time
Original EMF = 1.5v
Thanks very much!!
the plot of current and voltage would always have the same shape. They would be related by V(t) = I (t) * 1 ohm.
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It's this internal resistance that causes the voltage to drop when the current demand grows. This will decrease the actual available voltage from the
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You would need to know if open circuit voltage of the battery changes with time also to get the internal resistance, Rint. If you assume that the open circuit voltage, Voc is constant, then you could work this out.
V = Voc - I *Rint = V * Rint / (1 ohm)
Rint = (1 ohm) (Voc-V)/V
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How to find internal resistance using just voltage and current values?