PD or Iota for my upgrade?
I'm thinking about a converter and battery upgrade. The batteries will be wet cell lead acid 6 volt Possible T-105s or the 230AH, EGC2 from Sam's club. I'll probably go for four. (Does nayone have...
Forum Discussion
No, it doesn't work that way. The battery has just one impedance, not two as you suggest. The resistance will have minor variations over frequency, but we're discussing source resistance at very low frequency. You say it's so elementary that there are two distinct impedances that are orders of magnitude apart, then there should be many links describing this unusual feature.
Have you heard of Kirchoff's Voltage Law? If you do a loop analysis of the circuit (perfect voltage source plus series resistor connected to a battery charging source), you would realize the internal resistance of the battery can not be 1.25 ohms. According to KVL, the sum of the voltage drops of the voltage source and series resistance must equal 12.5V when 10A are conducting into the battery. At 10A, the internal resistor creates a voltage drop of 12.5V. That means the pure voltage source (battery voltage minus internal resistance) has zero volts! That's impossible. The actual battery source voltage is
V = V_at_bat_terminals - R_internal_resistance * I
V = 12.5V - 0.05 ohm * 10A
V = 12.0V
The actual battery voltage is 12.0V, not zero.
http://batteryuniversity.com/learn/article/how_to_measure_internal_resistance
Sal
Have you heard of Kirchoff's Voltage Law? If you do a loop analysis of the circuit (perfect voltage source plus series resistor connected to a battery charging source), you would realize the internal resistance of the battery can not be 1.25 ohms. According to KVL, the sum of the voltage drops of the voltage source and series resistance must equal 12.5V when 10A are conducting into the battery. At 10A, the internal resistor creates a voltage drop of 12.5V. That means the pure voltage source (battery voltage minus internal resistance) has zero volts! That's impossible. The actual battery source voltage is
V = V_at_bat_terminals - R_internal_resistance * I
V = 12.5V - 0.05 ohm * 10A
V = 12.0V
The actual battery voltage is 12.0V, not zero.
http://batteryuniversity.com/learn/article/how_to_measure_internal_resistance
Sal
DryCamper11 wrote:Salvo wrote:
What's the battery resistance when it is providing 10A at 12.5V, and 11A at 12.55A?
I realized I didn't answer this.By your method, it would be R = 12.5V / 10A = 1.2 ohm!!!
That's absurd. Given the data, the resistance is 50 m ohm.
It's not absurd. The battery looks like it has a DC resistance of
R = 12.5V / 10A = 1.2 ohm
when the converter is supplying 10A to a battery that has reached a charge of 12.5V.
It looks like a DC resistance of
R = 12.55V / 11A = 1.14 ohm if the the converter is supplying 11A to a battery that has reached a charge of 12.55V.
Those are the steady state impedance values for the two different steady state conditions. Those are the values one would use for testing the DC output capabilities of a converter. The dynamic impedance right around 12.5V at that charge level is as you say, 50 milliohm.
