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# Ohm's law is calling for a 50ohm resistor. Could I use two 100ohm resistors in parallel or would that be too close?

I'm trying to teach myself resistors through a LED project. My LED's typical FW voltage is 3.5 under a 4.5 supply voltage and a FW current of 20mA. If I've done the math right I should need a 50ohm resistor. correct?

I'm hoping to buy the resistors from Radioshack and looking at there supply they only have resistors ranging from 48ohm to 68ohm. I know I could just use the 68ohm resistor, but have also read that wiring resistors in parallel will reduce the resistance. So using two 100ohm combined should work. It just seems too simple and close to Ohm's law's outcome. Would this be safe for the LED?

To answer your question, yes 2 resistors 100 ohms each in parallel will be 50 ohms. It is that easy when they are equal.

To go further, most projects don't need exact values. You need to try to get close but any part you get will have some tollerance and not be exactly what it is labeled at.

You can get 1% resistors but they are only needed where highly accurate parts are needed like in measuring devices and frequency devices. They are a lot more expensive and harder to get. RS does not carry them.

Most devices are built using 10% tolerance resistors. So your 100 ohm resistor might be 90-110 ohms. But running an led that would be fine.

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Resistors in series ADD their value to each other.

Capacitors in PARALLEL add their value to each other.

most resistors are only 5% or 10% accurate anyway. That is

what the last color band means. GOLD is 5% Silver is 10%.

So you see they are OFTEN not extremely critical in value if you are

somewhere close within reason. Some circuits require PRECISION

resistances but not LED's. Here is a little trick to calculate ANY total

resistance of ANY number of resistances in PARALLEL. (if you have a

calculator with the (one over X) function which looks like this 1/X .

lets say you have 3 resistors you want to connect in parallel. their values are 1000 ohms, 1000 ohms, and 500 ohms. now grab your calculator.

punch in 1000............... now hit 1/x

hit the "plus" key

punch in 1000...(for 2nd resistor).... hit 1/x

hit the "plus" key

punch in 500 ..... (3rd resistor)..... hit 1/x

hit the equals key. then hit 1/x one last time

your answer should be 250

this will work for any number of resistors in parallel.

here is the equation you are using:

_________1_________

parallel OHMS total = 1/R1 + 1/R2 + 1/R3 +etc

Resistors in SERIES just add together.

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A normal LED will work fine with anything from about 10mA up to it's maximum rated current (If) which is usually 20mA or 25mA. If you run them a bit below maximum, you'll get very nearly full brightness and your batteries will last longer.

The forward voltage (Vf) you gave as 3.5V. This will be the maximum drop, and you'll probably find in practice that your diode gives less of a drop than that in the circuit.

Going back to your question, the answer is YES - If you've got two identical resistors in parallel, the resistance will be half (and the power capability will be double) but personally I'd put in a single 68R to allow for variations in Vf and to drop the current slightly.

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Whoops - I've just re-read the question and see you've given

typicalVf - normally Vf(max) is quoted. My mistake, but I'd still use 68R.Select as Best AnswerUndo Best Answer