Hi
@Boots in Action,
I think your maths is incorrect. P = V*V/R. Therefore R = V*V/P. In your example, V = 12V and P = 160W. Feeding that into a calculator give a resistance of 0.9 Ohms.
Quick and dirty method: 1 Ohm across 12V would draw 12A. Power consumed would be 12A * 12V = 144W. So 160W would need a lower resistance. The difference of 144W to 160W is about 10%, so reduce 1 Ohm to 0.9 Ohm would dissipate approximately 160W.
So, one would expect a reading of about 0.9 Ohm across a good 12V 160W element, not 13-15 Ohms.
cheers
Mike