A lower voltage may not affect your electronics' performance, but you should always check with your device's user manual to see if a battery voltage or chemistry is compatible. Yes, rechargeable D batteries use NiMH (nickel-metal hydride) or LiPoly (lithium-polymer) chemistries and have an 8.4V voltage compared to the typical 9V of primary variants. At least, it is so long as the current is low enough that the reaction rates can sustain it.Frequently Asked Questions Are there rechargeable 9V batteries? The battery voltage is held up in this way. When you place the battery in service and the accumulated charges disperse rapidly into a circuit to motivate a current to move (very rapid process that is only very rarely discussed in electronics), current in the circuit occurs and charges at the two battery terminals will recombine and this will disturb the field gradient inside the battery, lowering it, and thus the equilibrium state shifts out of that equilibrium with a renewed, slight preference for one side of the chemical equation, again, and electrons are moved through the inside of the battery until things are back in equilibrium. This sets the unloaded battery voltage that you'd measure using a high impedance device designed for that purpose. The increasing electric field gradient increasingly retards the chemical reaction rate, too.Įventually, the early preference for one side of the balanced reversible chemical reaction for moving electrons from the cathode to the anode has diminished enough so that the other side of the balanced chemical reaction occurs at the same rate and the battery finds an equilibrium state. This continues for a while and as it does an increasing field gradient (electric in nature) develops because of the accumulating charge differences. When all the materials are first assembled (for example, the diluted sulfuric acid is first poured into a lead-acid wet cell), the reversible chemical reactions inside the battery cell have a preference to start removing electrons from the cathode and adding electrons to the anode. Suppose you fabricate a chemical battery, using any of a variety of practical arrangements. (I'll also talk about a rechargeable battery case, with reversible reactions - although that's not critical to what I write below, it makes it a little easier for me to write.) And keep in mind that the usage of anode and cathode as a chemist applies them may at first seem confusing to someone in electronics. I'll just talk generally about the chemistry. Hence, how does a 9V battery generate 9V for an extended period of time? Obviously reality doesn't match my expectations but the saying "voltage is pressure current is flow" doesn't really explain this situation either. I conclude the potential difference must be less than 9V significantly. If electrons are flowing from the negative terminal through the light bulb and back to the positive terminal then how is it possible to have a 9V voltage difference between terminals for the entire hour? After 30 minutes wouldn't I be left with 4.5V worth of electrons? Half of my electrons have flowed from the negative terminal to the positive terminal. In one hour the battery runs out and the light bulb turns off. So lets say I connect a light bulb that consumes 500 mA at 9V. In a 9V battery the manufacturer might label the battery has 500 mAh (it can produce 500 mA for one hour.) I conclude there are enough electrons in this battery to create 500 mA for 1 hour. Voltage is a difference in the amount of electrons between 2 points. We can quantify the number of electrons moving past a point into a measurement called current. Electrons moving is what we call electricity. I recognize that all things are made up of atoms. In my career I have mostly ignored this question and just taken it for granted that if a 9V battery says it produces 9V that it will produce 9V until it runs out of "battery" but curiosity finally caught up with me and I want more details.
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