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Why we take copper losses I^2 R and why not I^2 Z? When we are measuring copper losses in transmission lines or generator motor or transformer we use above case why we do so?

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    \$\begingroup\$ Hi and welcome to EE stack exchange! Please tell us a bit more on the 'background' of your question. Maybe draw a circuit of your problem to show us what you are thinking. \$\endgroup\$ Commented Oct 16, 2018 at 4:37
  • \$\begingroup\$ specifically, why you think you should be using Z instead of R, and what Z is. As Stefan said, we really don't know how to help you without a little more background. \$\endgroup\$ Commented Oct 16, 2018 at 5:31

1 Answer 1

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If Z is the combined resistance and reactance, only the resistive part R will lead to losses. The reactive part only consumes reactive power

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  • \$\begingroup\$ But reactive is also loss then why nit consider reactive loss only resistive loss is considered why? \$\endgroup\$
    – KHAAN
    Commented Oct 17, 2018 at 9:19

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