Attach a 12V fan to this circuit which draws 70mA (0. 007A), so the circuit must have at least that much current. Yes, there is something missing in the analysis. Actually there is too much in the analysis. The voltage drop across the resistor will be simply be the voltage you are putting across it, i. e. 12. 6V (or 12, 6 in your numeric language :) angelatlarge Apr 7 `13 at 4:30
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The issue here is probably coming from the term "voltage drop". If you measure 10, 83 V at the resistor then your are either losing voltage to your cabling(probably not at 1ma) or you have a battery that is very near end of life and sagging a little. Your voltage "drop" on the resistor is how much the voltage changes across the resistor, not how much it changes getting their. So 10, 83/10k is. 00183 A or 1. 83 mA, which is reasonable for a 10k resistor. Kortuk™ Apr 7 `13 at 4:47 so, for 1, 77V drop, that is happening in your cabling or battery, so if you take that drop with the current you re calculating you can determine the resistance of the cabling+series resistance of battery that is being placed with R1. I would write an answer, but someone with pictures would make a much better job then me. Kortuk™ Apr 7 `13 at 4:54 @angelatlarge, you are right. the voltage across the resistor will be the same as there is only one load, so the current will be I = V/R, I = 12. 6V / 10K = 1. 26mA. David Norman Apr 7 `13 at 5:19 @vsams14 I believe you might have some misconceptions about what voltage and current from a power supply are. this question is very detailed and might help you. If you are just asking if your power supply will work for this fan, if the fan is designed for a 12V supply you are 95% fine. Kortuk™ Apr 7 `13 at 6:25 LEDs are current-based devices, not voltage based. In other words, the design parameter for a driving...
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