When I connect the positive and negative leads of an ohmmeter to pin 1 and middle pin of the SoftPot I get an accurate reading from the softpot (that is 0 to 10 k if I remember correctly)
For 5V this would be 100mA which may indeed burn the sensor if it expects 7mA only.
Assuming that the softpot is 10K as advertised, putting 5V across it will result in 0.0005 amps (0.5 mA) which equals 0.0025 watts (2.5 mW)
With a supply voltage of 5V, power would be as low as 2.5mW for a 10k resistor
.. but this is a passive linear component and will operate at 0.5W with a variate of currents and voltages ..
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