Rate of energy dissipation in a wire carrying current cut in half [closed]

Question

In my physics textbook, I have a sample problem stated as follows:

You are given a length of Nichrome heating wire; it has a resistance R of 72 Ω. At what rate is energy dissipated in each of the following situations: (1) 120 V is applied across the wire. (2) The wire is cut in two halves, and 120 V is applied through one half.

(1) $$P=\frac{V^2}{R}=\frac{(120 V)^2}{72Ω}=200W$$ (2) $$P'=\frac{V^2}{R'}=\frac{(120 V)^2}{36Ω}=400W$$ and for the two halves, $P=2P'=800W$

This is 4x the initial dissipation rate. Thus, you might conclude that you could buy a heating coil, cut it in half, and reconnect it to obtain four times the heat output. Why is this unwise? (What would happen to the amount of current in the coil?)

There is the fact that maybe that a bigger current would be too much for the wire, but I don't think that that is the solution. But I don't see why this would be unwise: the initial current being 1.66 A, then 3.33 A for one half, when both halves are connected in parallel, they will each receive 3.33 A. How is this a problem?

Answer 1

A heater wire is in a state of equilibrium between the rate at which power is supplied and the rate at which the heat is radiated snd conducted away. If you double the power and half the surface area, the temperature will go way up. If the length of the original wire was chosen to give a functional temperature then running it up is likely to melt the wire.