What Is the Correct Power Calculation for a High-Torque DC Motor at 8V?

[chunkytuna21]

A high-torque dc motor designed to drive cassette decks has a no-load speed of 7400 rpm at 9 V with a torque of 9.6 in.·oz and a no-load current of 22 mA. How much power will it draw from a 8 V battery when turning freely? [Hint: At what voltage is it operating and how much current does it draw?]

I= V/R
R= V/I
P= IV

22mA= 0.022A
R= 9V/0.022A= 409
I= 8V/409= 0.019559902
P=IV= (0.019559902)(8)= 0.16W

0.16 is incorrect. Where am I going wrong? Should I round up to 0.2? Is there something wrong with the problem? I am out of ideas.

 

[Andrew Mason]

A high-torque dc motor designed to drive cassette decks has a no-load speed of 7400 rpm at 9 V with a torque of 9.6 in.·oz

What is the work being done here? There is no load, but that does not mean that there is no friction. The motor does mechanical work with no load. If it did not, there would be no torque applied with the motor running at constant speed. The difference between the electrical power consumed and the rate of work being done gives you the power loss in the motor:V^2/R so that enables you to find the resistance. Take it from there.

AM