' post is entirely correct, it doesn't explain the 1.9 mA. Have a look at the datasheet of your multimeter, specifically at the current measurement section and there look for burden voltage or similar.
In this example
View attachment 53558
we see a max. burden voltage of 1 V in the 200 mA range, which translates to a series resistor of 5 Ω. This can have an influence on your measurement.
Still Using 20 Ω instead of the 15 Ω you claim to have used still doesn't explain the measurement:
7.3 V / 20 Ω = 365 mA, way off from your measurement.
Next look at the pass voltage of the LED. @Bluejets
assumed a typical red LED with ~1.7 V pass voltage. Assuming you used a white LED with a pass voltage of 3 V (r higher) we arrive at:
(9 V - 3 V) / 20 Ω = 300 mA.
None of these calculations explains your observation. I suggest:
- Show us a complete diagram of your circuit. It always helps to understand what's going on.
- Check your resistor. Is it really 15 Ω? The measurement suggest something in the kΩ range.
- Check the pass voltage of the LED. What do you actually measure?
- Check the battery voltage under load. What do you measure (with the LED + resistor connected)? the typical 9 V block tends to lose voltage quite rapidly.