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From the datasheet of a modem transformer we have:

  • Impedance, Primary: 600 Ω Typ. @ 1 kHz 1 Vrms*
  • Shunt Inductance Lp, Primary: 3.2 H min. @ 1 kHz
  • D.C. resistance @20°C: Primary = 82 Ω ± 15%

Also the impedance formula:

$$ Z= \sqrt{R^2+(XL-XC)^2} $$

How does 600 Ω result from 3.2 H and 82 Ω? (The XC is negligible.)

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  • \$\begingroup\$ You will get your 600Ω only if the load the Secondary with a 420Ω resistor. Look at the asterisk note. \$\endgroup\$
    – G36
    Sep 28, 2018 at 17:50

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How does 600 Ω result from 3.2 H and 82 Ω? (The XC is negligible.)

Think about what the spec says and read this below.

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