Electrician from Australia here. Just having an issue with a heater fan in a bathroom for a client. According to the diagram I have posted, I can wire a second fan up to this and it should all work.
The issue I am having is that because of the permanent supply to the unit, once that thermal switch goes above 55C it supplies power to both the fans via the 20W 330ohm resistor. By adding a second fan as drawn by the diagram, this also will be powered in an over current situation via the resistor.
The issue is, I don't believe the manufacturer has realised the implications of having a second fan fed from the ceramic resistor, and it heats up and blows the thermal fuse at 93C.
So what wattage resistor is required to power the 25W main fan in the unit, and a second 60W fan external to the unit?
I tried working it out as below;
Current draw of main fan = P/V = 25/230V = 0.1A
Voltage drop over resistor= I*R= 0.1*330= 33V
Power dissapated by resistor with main fan= V*I= 33*0.1 = 3.3W
With the second fan of 60W in the circuit, we now have 85W load.
Current draw of main fan = P/V = 85/230V = 0.37A
Voltage drop over resistor= I*R= 0.37*330= 122.1V
Power dissapated by resistor with main fan= V*I= 122.1*0.37 = 45.17W
45.17W on a 20W resistor- Houston we have a problem.
Is my maths and methodology correct??