If you have two heating elements, one rated for 12V, and the other rated for 120V, _both_ rated for 5A, then the resistance _must_ be different.
power dissipated is P = I^2*R (square of current times resistance)
power dissipated is also P = I*E (voltage times current)
voltage across a resistor is E = I * R (voltage = current times resistance)
notice that you can substitute E = I * R into P = I * E to get P = I * I * R, or the first equation.
In your example, the 12V 5A resistive heating element must have a resistance of 2.4 ohms, and must dissipate 60 watts. The 8*10^10V 5A heating element must have a resistance of 1.6*10^10 ohms, and must dissipate 4*10^11W (400 nuclear plants worth of output *grin* )
The confusing bit here is that E the voltage in question is _not_ the total voltage of the system, but instead the voltage dropped across the element being considered. With the fuses mentioned above, the supply voltage is being dropped across the _load_; and very little voltage is being lost in the fuse itself.
If you have a 250V 10A rated fuse, you could connect that fuse in a 120V circuit with a 500W lamp. The resistance of the fuse might be 0.01 ohms, and the voltage drop across the fuse might be 0.04V, with a total power dissipation in the fuse of 0.17W. The voltage drop across the lamp is 119.96V, and nearly 500W is delivered. The resistance of the lamp would be about 29 ohms.
Take this same fuse and put in in a 12V circuit with a 50W lamp. The current flow is still 4.2A, and the voltage drop across the fuse is still 0.04V with a power of 0.17W. But now the voltage drop across the lamp is 11.96V, and the resistance of the lamp is 2.9 ohms.
-Jon
