Calculating the resistors value?

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SUMMARY

The resistor value can be calculated using Ohm's Law, where R = V/I. In this scenario, the supply voltage (Vs) is 9 volts, the LED voltage drop (Vled) is 3 volts, and the LED current (Iled) is 10mA. The voltage across the resistor is determined by subtracting the LED voltage drop from the supply voltage, resulting in a voltage of 6 volts across the resistor. Therefore, the correct calculation is R = 6V / 0.01A, yielding a resistor value of 600 ohms.

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Calculate the value of the resistor using the following information:
Vs=9volts, Vled=3volts, and Iled=10mA


I'm assuming that it's just Ohm's Law R=V/I,

but I don't know which of the two voltages I use, though I think its the Vs...but I don't know what the 's' stands for either.
this is a pre-lab question, and I should probably be able to do it but I've been sick all (last) week so have missed all lectures relating to this...



R=V/I
= 9/0.001
R=9000ohms

this seems like a very lrge number, and also too easy for a pre-lab, they're usually very difficult.

Thankyou for any help offered. :)
 
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I'm guessing that the circuit is set up with a voltage source (Vs), an LED with voltage drop Vled and a resistor in series. So you need to work out the resistance of the resistor such that it takes the supply voltage and reduces it to the voltage across the LED.

Does that make sense?
 
it does make sense, but I don't know how to do that... or is that what I've done?
Is the Vled value there just as extra information that isn't needed or?!?...
 
Kirchhoff's loop rule says that the net voltage in a circuit is zero; i.e., if the source generates a voltage V, then there is a voltage drop V across the circuit. You already know that the LED "reduces" the voltage by 3 V, and you know the voltage generated by Vs(the source or battery or whatever).

From that you can find the voltage drop that MUST be across the resistor, and you know the current. Then just use Ohm's law.
 
Also be careful with where you put your decimal point for the current. 10mA is not 0.001A it is
.01A.
 

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