Since we don't know the voltage across the reactor, we can't use P=V^2/R.
Thankfully, we know the value for current, so we can use P= R*I^2
Now, R is in the numerator and we can't say that assuming R=0 makes for a conservative measurement of power.
This isn't voodoo trickery, it's just a consequence of V=R*I; assuming R=0 is the same as assuming V=0 therefore it is not a conservative way to measure power.
The only case it would be conservative is if we had a measurement of V(reactor+resistance). Then we can use V^2/R and assume R(reactor)=0.