Battery Charger Question

Besides temporarily disconnecting the charger, they may also be comparing the charging current to the battery charge profile. See if this link helps.

In or on the box it came in, is a piece of documentation called the "instructions". Somewhere on the box, or on this document, is the name of the manufacturer. A few seconds' work will locate the manufacturer's telephone number. After this, a training package called "How to use the telephone 1.0" needs to be undertaken, after which the equipment manufacturer can respond to this and all other technical queries on the equipment. Go for it.

Don't make it more complicated, as it is plain and simple: Battery voltage depends on state of charge. Charger output voltage depends on whether it "sees" an open circuit, a short circuit, or something in between (like a battery that is not charged up well).

When you check battery voltage, the charger is temporarily out of the circuit.

Charging amps is a completely different story, and the charger is supposed to be sophisticated enough to know not to "blast" 100 amps into a spent battery, so it does not warp the plates (excess heating of the battery can warp parts).

There you go.

There is a current limiting resistor in the circuit between the battery and the charging circuit. The switch selects on which side of the current limiting resistor the voltage is read. The voltage drop across the limiting resistor is determined by the charge current and the indicated battery voltage will be just very slightly higher than the actual battery voltage, but not enough to be really significant.

Is the charge current limiter about the same value as a current shunt? Or is it more like 1-10 Ω?

If the charging current is limited to 10 amps and the supply voltage is 15V, then the current limiter will be around 1.5 ohms. More likely it is a current regulator that will vary from 1.5 ohms down to .2 ohms as the battery voltage rises from zero to 13 volts.

If charger and battery are wired together their voltages are the same. No voltmeter can tell different, because the voltages are the same (excluding possible wiring resistance volt drop due to charging current).

AS others comment, the "voltmeter" switch may be doing more - like turning off AC power to charger, or opening DC circuit from charger to battery.

The battery voltage will not instantly fall when it is taken off charge, especially if there is no load on the battery. I was told by a battery maker that it takes 24 hours for an open circuit battery to reach a true open circuit voltage. Usually, a load is applied briefly to speed-up getting a stable voltage.

The open circuit voltage of a lead-acid battery is 0.84 + specific gravity.

e.g. with S.G. = 1.21, open circuit volts = 0.84 + 1.21 = 2.05 volts per cell at 15 celsius.

There is no such relation for an alkaline battery.

Good points.

I wish I knew more about this IFBattery from Purdue Univ. Which of the six or seven candidate chemistries will win out?

What will be the operating parameters of that chemistry, or will it be:

"My good solution tank is 1/4 full, and my spent tank is 3/4 full".

"Siri - find me an electrolyte station", or they might as well call it a redox station.

OR, it could be a gas station, and a redox station.

Sorry, you're wrong.

If the battery and charging circuit were always the same voltage, then there would be no current flow and no charging. To charge, the battery must always be at a lower voltage than the supply.

For their voltages to be the same "wired" together, it would have to be a hefty shunt, with a very, exceedingly low resistance. Normal wiring would heat up, and then heat up some more, leading to a very noticeable voltage difference, maybe even a battery heat up from dissipating too much current through the cells too quickly, since not even this electrochemical process is all free energy, some of it goes to TΔS heat. Too much a heata, the batteryie, she go booma!

I go with the point of the original post and the responses which say the switch or circuit is more than the label states....

In practical use, you could connect charger and battery with very thick or even superconducting cables, so the difference is unseeable on the voltmeter. The charger must have a higher emf than the battery, but the difference is hidden inside the charger in its impedances when they are connected and a current flows. The difference in voltage tells one nothing unless you know the cable resistance or current - if you do the the calculation it will just tell you the cable is adequately low resistance.

If you apply a fixed voltage to a resistor, the current increases until the resistor terminal voltage equals the supply, in practice the connecting cable is made low enough resistance for the circuit to do its job.