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No, they are in parallel so the resistance should be 25 Ohms. Also, there are two 1/4W resistors so, you should use a 1/2W resistor.
Perfect! Thank you very much!
So, with that in mind, does this mean i could put a 10-15 Ohm 1W resistor on 4 LEDs?
Let's do the math... R = 1/( 1/50 +1/50 + 1/50 +1/50)
=1/( .02+.02+.02+.02) = 1/(.08) = 12.5 Ohms
so, you could replace all four resistors with a 12.5 Ohm resistor.
Power would be 1 * 1/4W = 1W
So, you could use one 12.5 Ohm 1W resistor to replace all 4
CORRECTION: Power would be 4* 1/4W = 1W (I got ahead of myself went typing the above)
If you have really no other option... you may try. If x is the number of parallel diodes: 50Ohm/x, 0.25W*x.
BUT: You should not do that. LEDs have minor differences. If you put them on a shared resistor, one with (the slightly lower Uf) will get more current than the other(s). The results may vary from not even being noticeable to a thermal run away and break down of the LEDs.
So it may work but is not recommended.
No, you need to use 50Ohm resistors. Or you need 5 X 10 Ohm resistors, which can be 1/5th Watt.
FYI, to calculate the total resistance of paralleled resistors, the formula is:
When putting resistors in series, you just add the values to get the total resistance. When putting resistance is parallel, you add the CONDUCTANCE (1/R) and then convert back to resistance.
If they are the same ...... Just divide by the number of them !
Yes, but I wanted to show the proper way of calculating them so when he runs into different values...
And you did it very well.
Fifty years after graduation, the number of times Iv had to work uneven parallel resistors is counted on one hand.....
Paralleling LEDs taken to the extreme just one resistor used here see the brightness variations on hand test selected LEDs !!!!
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Posted:Jan 7, 2015
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