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I am using a simple 12v dc motor circuit with the following items wired into it. I have tried to provide as much information as possible. So far the circuit works great but I want to be able to adjust the motor speed. The motor is on a H-bridge type relay to allow me to reverse the direction. 12vDC lead acid battery Motor 12v - power rating 6.21w relay 1 12v latching relay (max voltage 220v) 100mΩ contact resistance relay 2 max DC 240v maximum current 10A contact resistance 50mΩ I want to add a potentiometer/variable resistor to the circuit to ramp the voltage between 9v and 12v to adjust the motor speed. I have tried a few pots but keep burning them out due to my lack of voltage/current/resistance knowledge. Can someone suggest a simple way of doing this and show me the working so I can work it out myself in future. No 'see a proper electrician' comments please - not helpful! Thanks in advance. Any rating details I should look for on my pot would be great. Thanks!
And got the following answer:
If the motor is rated 6.21W at 12V, the rated current is 6.21/12 or about 0.5A. At 9V it could still draw 0.5A depending on what it is driving. If you want to reduce the voltage by 3V at 0.5A, the resistance should be 3/0.5 or 6 ohms. The potentiometer or variable resistor should be able to carry 0.5 amps. If you use a 10 ohm resistor, it would need to be rated for at least 10 ohms X 0.5A squared or 2.5W (5W would be better). If you use a 25 ohm resistor it will need to be rated for at least 6.25W (10W would be better). If you calculate the wattage of the variable resistor using the maximum current and add a safety factor, the resistor should be able to carry that current as a variable resistor regardless of what resistance it is set for. If you connect it as a potentiometer, you will need a higher power rating because one end of it will need to carry current to both the motor and the other end of the potentiometer.