Quick question about voltage (DC) - diyAudio
 Quick question about voltage (DC)
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 18th November 2006, 07:05 AM #1 diyAudio Member   Join Date: Nov 2006 Quick question about voltage (DC) Oh boy... I didn't realize just how little i know about electronics till I thought to finally ask someone about this... You guys are gonna cringe...lol Lets say we have a flashlight... IF: [1.5v battery] connected to [BULB] - Gives you 1.5v worth of light, over the lifetime of the battery... let's say 10 hours just for example. THEN, DOES: [1.5v battery]+[1.5v battery] = 3.0v over 10 hours, OR [1.5v battery]+[1.5v battery] = 1.5v over 20 hours?
 18th November 2006, 07:42 AM #2 diyAudio Member     Join Date: Jan 2005 Location: Phoenix, Az. It depends on the type of battery, but they all lose voltage as time passes. Some lose it slowly then it suddenly drops, others lose more or less continuously. Alkaline cells lose voltage sort of continuously, but recover a bit if the load is removed for a while. I_F
 18th November 2006, 04:48 PM #3 diyAudio Member     Join Date: Jun 2002 Location: USA, MN Actually, pretty much neither scenario is correct. Perhaps a better way to phrase the question would be: If you use a 1.5V bulb and a 1.5V battery and it lasts 10 hours what happens if you use 2 batteries either in series to get 3V or in parallel to get 1.5V? If you put the batteries in series, you will get 3V, which will make a brighter light that would probably last about the same length of time, perhaps a bit longer. If you measure the current, you might get a better idea. If you put the batteries in parallel, they will share the load equally (assuming they both have the same internal resistance) and the result would be a battery pack that lasts for somewhat more than 20 hours. This is because the energy you get out of a battery depends on discharge rate. Lower discharge rates give more total energy output. Higher discharge rates give less. A battery rated at 200maH might last 200 hours on a current drain of 1ma, but it wouldn't last for 1 hour with a current drain of 200ma. A more exact answer would depend on the type of battery, and then you would have to measure the battery performance curves because they aren't generally available and come up with a battery "model" or a mathematicla description that predicts battery performance. You would then need to make a model of the bulb, because its resistance and brightness changes with the voltage across it. __________________ Our species needs, and deserves, a citizenry with minds wide awake and a basic understanding of how the world works. --Carl Sagan Armaments, universal debt, and planned obsolescence--those are the three pillars of Western prosperity. —Aldous Huxley
 19th November 2006, 03:48 AM #4 diyAudio Member     Join Date: Aug 2006 Location: Texas Blog Entries: 2 Okay, let's cover this without all the real-world parasitic horrors first. In a flashlight, the batteries are put in a series configuration. As they are end-to-end, one comes right after the other - hence, "series". In a series combination, the voltages add - 1.5V+1.5V=3.0V. If there is twice the voltage, then twice the current is inevitable - ohms law: V/R=I (I is current). So 3.0V/R=2*(1.5/R). If there is twice the current, then the battery will run out faster than with only one battery (only 1.5V). If the batteries are used in parallel, then the voltages don't add. However, since there are two batteries, the current draw is split between them. So if the lamp draws 2mA, and the battery rating is 200mAH, then one battery will last 100 hours. However, since we are using two batteries, the current is split between them - hence, 1mA from each battery. This causes the pack to last twice as long. Does that clear things up? -keantoken __________________ The Kmultiplier rail filter! -=- The Simple Kuartlotron Superbuffer! Need something built, repaired or modded? PM me and ask what I can do!
 19th November 2006, 09:41 AM #5 diyAudio Member   Join Date: Mar 2006 Location: nsw Or to look at it in a practical setting, lets assume two separate torches have been designed. One to use two cells in series, and one to use them in parallel. They are each designed to put out the same amount of light and they will both last the same amount of time. To explain this, one offers three volts. The hypothetical bulb would have a resistance of 6 ohms. A half amp would flow and there would be 1.5 watts delivered to the bulb. In the second example, there is 1.5V. The bulb has a resistance of 1.5 ohms. One amp would flow (half an amp per cell) and there would be again 1.5 watts delivered to the bulb. In both cases the batteries pass 500mA.
 19th November 2006, 09:57 AM #6 diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders Hi, The series (3V) bulb will have finer wire than the single cell bulb. That makes the single cell bulb more robust against vibration and the bulb will last longer when starting from the thick wire before evaporation erodes it. The parallel cells layout delivering the same power should last a tiny bit longer provided the the resistive losses in the two arrangements are the same. (series can use lighter metal components). I would have expected manufacturers to have jumped on this long before now, but they have not, so there must be a flaw in the argument. Is it weight or cost or some other? __________________ regards Andrew T. Sent from my desktop computer using a keyboard

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