Given a current transformer like this:

typical current transformer

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.

  • \$\begingroup\$ Reflected impedance in a transformer goes as the square of the number of turns. \$\endgroup\$
    – John D
    Commented Aug 10, 2015 at 23:07

1 Answer 1


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×.

  • \$\begingroup\$ Interesting, but for most current transformers, like the one above, the # of turns is not given. At most, the output voltage for a given input current is. Is that sufficient for calculating the effective insertion resistance? \$\endgroup\$
    – iwantmyphd
    Commented Aug 11, 2015 at 0:54
  • \$\begingroup\$ I think you're confusing two current-measuring technologies. Current-sensing resistors are usually specified in terms of voltage drop for a given maximum current, which is really just another way of specifying their resistance. Current-sensing transformers are always specified in terms of their turns ratios, along with their maximum current rating. \$\endgroup\$
    – Dave Tweed
    Commented Aug 11, 2015 at 1:03
  • 1
    \$\begingroup\$ ... but in any case, in terms of your specific question, the exact number of turns on the secondary doesn't matter -- regardless of what it is, if you have 10x the turns in the primary and keep everything else the same, the primary impedance will go up by 100x. \$\endgroup\$
    – Dave Tweed
    Commented Aug 11, 2015 at 1:08
  • \$\begingroup\$ thanks, I'm just going off of the mfg's website. They refer to insertion resistance when using their current transformer. Here's the link: pearsonelectronics.com/insertion-resistance \$\endgroup\$
    – iwantmyphd
    Commented Aug 11, 2015 at 1:48
  • \$\begingroup\$ What about the DCR of the secondary and the burden resistance .this also gets reflected \$\endgroup\$
    – Autistic
    Commented Oct 22, 2015 at 2:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.