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I don't understand how I am "calculating for the voltage drop"by BeautifulLight
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#1
Sep114, 09:56 AM

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Assume I have connected a single lightemitting diode (2V, 30mA) and resistor to a standard 9V battery. Ohm's Law is used to calculate the value for the resistor. The observed relationship between voltage, current, and resistance is as follows, V=IR.
I intuitively understand there will be a voltage drop across the diode because it takes a certain amount of energy (in joules) for electrons to "jump" from cathode to anode, and voltage is defined as joules/coulomb. Therefore, the potential difference will now be 92=7V. So, 7=.03R, R=233Ω What's the problem? I understand everything I just did, and yet I am oblivious to the fact that when I solved for value R, I was also solving for a 7V drop across the resistor. Would it be better to disregard the diode and just use a single resistor to help me understand? So, 9=I(233). I randomly chose that value for R. I'm not sure how you figure voltage drop using only 9=I(233). From that, I can only derive current, nothing else... 


#2
Sep114, 11:21 AM

P: 410

If you put your 233 ohm resistor across a 9 volt battery, it drops 9 volts. Voltage drop = voltage across resistor.
Resistor "see" 9V across his terminals because you connected a 9V battery directly across a single resistor. So in this case resistor will only change the current in the circuit. Because how can a resistor change the battery voltage? We have a different situation when we add another resistor in series 


#3
Sep114, 11:23 AM

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EDIT  Nice post by Jony! 


#4
Sep114, 12:52 PM

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I don't understand how I am "calculating for the voltage drop"
Remember beginning algebra  one equation per unknown? If A = B X C, you must know two things to figure out the third one. Ditto for V = I X R . Then you wrote in .03 for I. You got .03 from LED datasheet. So you were left with one equation and one unknown, which is solvable. And you solved it. Keep it simple. Talk your way through these things  talking stimulates different parts of the brain than does just that "inner voice".. 


#5
Sep114, 05:48 PM

P: 14

I thought voltage drop was strictly the result of a component actively using voltage. Remember this?
"I intuitively understand there will be a voltage drop across the diode because it takes a certain amount of energy (in joules) for electrons to "jump" from cathode to anode, and voltage is defined as joules/coulomb. Therefore, the potential difference will now be 92=7V." So when you told me there was a 7V drop across the resistor, I immediately asked myself how in the world a resistor "uses" 7V and hence the last part of my initial post, "Would it be better to disregard the diode and just use a single resistor to help me understand? So, 9=I(233). I randomly chose that value for R. I'm not sure how you figure voltage drop using only 9=I(233). From that, I can only derive current, nothing else..." ***There is a difference between voltage drop as a result of a component actively using voltage VS voltage drop by simply shorting out the leads of a 9V of course the drop is going to be 9V, that's the EMF of the battery. In my case, it was easiest for me to think of the 7V as "leftover" voltage ...having absolutely nothing to do with the value of the resistor. I have one additional relevant question, Why should a lightemitting diode have a minimum current rating such as 30mA? As long as there is enough voltage supplied so electrons can jump from cathode to anode (or vise versa), then there should be no reason why electrons shouldn't flow even if there's only 0.00000000000001 amps supplied. Is the 30mA some kind of threshold for the naked eye? 


#6
Sep114, 05:59 PM

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As you learn more about basic electronics, you will stop using terms like "voltage used up" and so on. Just thing about voltage drops due to PN junctions, resistors, and so on. Later you will learn about reactive elements like inductors and capacitors, which can have AC voltage drops across them that vary with frequency... 


#7
Sep114, 06:28 PM

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#8
Sep114, 06:51 PM

P: 14

I may be very wrong, but I'm letting you know how I see it. Edit: For instance, maybe the resistor does require a tenth of a volt for current to pass through it, but that figure is so insignificant compared to the 7V that has been leftover. If that figure would be more significant, then we would have to start dropping voltages on every resistor as we did with the 2V drop on the lightemitting diode. I thought about this^ over dinner and it's (partially) incorrect, but mostly! That 233Ω resistor IS using all 7V granted there's .03 amperes of current. I forgot about Ohm's Law... I say partially because when I read your post, it made it seem like or at least to me, that all 233Ω resistors require 7V. That isn't true. You can double your power (14V, .06 amperes) and it still works out, hence 14=(.06)(233) or 14=14. 


#9
Sep114, 07:20 PM

P: 535

V = IR, or water pressure = flow*(1/diameter). Think about it, if you constrict the hose with constant pressure the flow increases (you use this to blast suds off your car, for instance). If you imagined a fat hose with ten times the diameter the water would just trickle out, since the volume of water (or the "power") would be constant. If you don't like the water analogy, ignore it (plenty of people hate it because it can be misleading at times).... in that case remember that for resistor circuits V = IR. Manipulate this equation to see how volts, amps, and ohms are related when it comes to flow and power. As you can see in the equation, the idea of "left over" volts makes no sense. All the voltage is dropped across the total resistance. 


#10
Sep114, 07:36 PM

P: 14

Thank you analogdesign. You caught me while I was editing my post (#8). Here
I thought about this over dinner and it's (partially) incorrect, but mostly! That 233Ω resistor IS using all 7V granted there's .03 amperes of current. I forgot about Ohm's Law... I say partially because when I read your post, it made it seem like or at least to me, that all 233Ω resistors require 7V. That isn't true. You can double your power (14V, .06 amperes) and it still works out, hence 14=(.06)(233) or 14=14. The only time I like to see water analogies is with "valves" (diodes/triodes) because you can take it literally Oh, and I don't mean to knock the whole insulator/conductor thing; I'm just one of those 0.0000000001 isn't 0 kinna guys. 


#11
Sep114, 07:40 PM

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Yes
I was in the middle of drawing a pic when you replied .... seeing a circuit often helps visualising the problem Dave 


#12
Sep114, 07:47 PM

P: 14

How did you make that schematic? I don't see anything underneath "reply." Is that another program you're using and you're just hitting the paperclip icon to copy it?



#13
Sep114, 07:48 PM

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so we know there is 9V across the resistor and LED ... ie. the voltage measured between A and C = 9V
the datasheet tells us that there will be a 2V drop across the LED ... between B and C that means there must be a 7V drop across the resistor ... between A and B BECAUSE all the voltage drops must add up to the total voltage applied to the circuit Dave 


#14
Sep114, 07:48 PM

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#15
Sep114, 07:48 PM

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If you click on the Go Advanced button below the text box, it gives you the option to manage attachments you can use that to upload files from your puter 


#16
Sep114, 08:02 PM

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so some things you can now work out
here's a problem for you to solve ... in a series DC circuit, the current is measured the same everywhere, therefore there is 30mA through the LED, through the resistor and through the wires coming from and to the battery Using Ohms Law  R = V/I knowing that the voltage drop across the LED is 2V and the current is 30mA  What is the resistance of the LED ? knowing that the voltage drop across the resistor is 7V and the current is 30mA  What is the resistance of the resistor ? Dave 


#17
Sep114, 08:09 PM

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