Given a current transformer like this:
there is an insertion resistance that appears as a very small equivalent series resistance on the measured circuit. Since it is usually small (say, 0.0001 \$\Omega\$), it can be neglected.
If you were to feed the current-carrying wire you are trying to monitor through the transformer multiple times to amplify your signal, does this also multiply the insertion resistance by the same amount? If I loop the wire through 10 times, does this multiply the insertion resistance by a factor of 10?
To clarify what I am asking, here is a link describing what insertion resistance means in the context of the type of transformer shown above.
Best Answer
As JohnD hinted, the impedance ratio of a transformer is equal to the square of the turns ratio.
So if you have a 1000:1 transformer with a burden resistance of 100 ohms, the "reflected" resistance on the primary will be
$$\frac{100 \Omega}{1000^2} = 0.1 m\Omega$$
If you change the turns ratio to 1000:10 (= 100:1), but keep the burden resistance the same, the reflected impedance becomes:
$$\frac{100 \Omega}{100^2} = 10 m\Omega$$
... which is 100× as large, not 10×.