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This question has been on my mind for some time. The best way I can try to explain this is by giving you a senerio:
When I apply 50% duty cycle of the PWM to a motor that has a load against it, the initial current draw is high, then comes back down to a steady state level. In this case the high current spike is at 10Amps, and the Voltage setting is at 12VDC. How can you get such a high current draw if the motor has terminal resistance of 1.5 Ohms? Using Ohm's law: V=IR, so I=V/R, the current can be no greater than 8 amps (12V/1.5Ohms), but I see 10 amps. Also, it should be alot lower since I only applied 50% duty cycle. Doesn't 50% duty cycle mean 50% of 12VDC, which is 6VDC? Am I thinking about this wrong?
When I apply 50% duty cycle of the PWM to a motor that has a load against it, the initial current draw is high, then comes back down to a steady state level. In this case the high current spike is at 10Amps, and the Voltage setting is at 12VDC. How can you get such a high current draw if the motor has terminal resistance of 1.5 Ohms? Using Ohm's law: V=IR, so I=V/R, the current can be no greater than 8 amps (12V/1.5Ohms), but I see 10 amps. Also, it should be alot lower since I only applied 50% duty cycle. Doesn't 50% duty cycle mean 50% of 12VDC, which is 6VDC? Am I thinking about this wrong?