eight 47 Ω resistors in parallel (5,87 Ω). As I = U / R (current is equal to voltage divided by resistance) ˜4 V / 5,87 Ω equals to 0,69 Amperes. Each of my eight resistors has to deal with a current of 86 mA and should therefore support 0.5 W (P = 0,086 A * 4,0 V = 0,35 W).?? Can you tell me what Resistors Part number I would need to do this.? Thank you michael
Hello @mphundley
Without knowing if you have any other requirements here is a link to 47 Ohm and 1/2W or higher through hole options.
https://www.digikey.com/short/p23hwc
-Robert
Robert. I want to Discharge Batterie cell manually that are with higher cell. I wanted to stay below one ampere and simply adding eight 47 ohm resistor. I don’t no what all that is can you let me no what resistor part number can do this. Michael
With total resistance, your wattage does not need to increase because you are still using 47 ohms per parallel branch. The overall current in the whole circuit would be .690 Amps or 690mA ; each resistor does not need to handle that. You take the voltage divided by the resistance 4 / 47 = .085 A per resistor. This means that each resistor would be needed to be rated for 0.34 Watts if you kept the parallel configuration. If you replaced the whole thing with a 5.87 Ohm resistor, then the one resistor would be needed to be rated for 2.72 W as Robert stated.
Sorry @mphundley I missed the error in calculations that @Kaleb_Kohlhase pointed out. Needing 690mA per resistor would be approx. 2.76W each. That would mean you would need a 3W resistor. 47AECT-ND would be an example of this specification. Here are some additional options if you want a different composition or want to go slightly higher on the wattage. https://www.digikey.com/short/p2qwvv