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Assume I have connected a single light-emitting 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

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...

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 9-2=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...

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