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As i know, we find a thevenin equivalent resistance by finding open-circuit voltage and short-circuit current and then divide these two. I have looked on wikipedia and other sources but could not find an answer. On these websites they only tell how to do it. How opening a circuit and shorting a circuit helps us to find an equivalent resistance? What is a physical fact behind this idea?

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    \$\begingroup\$ See this. Go to the bottom of the page and read up a little. \$\endgroup\$
    – jonk
    Apr 8, 2020 at 6:53
  • \$\begingroup\$ Maybe this explanation will help you electronics.stackexchange.com/questions/377467/… \$\endgroup\$
    – G36
    Apr 8, 2020 at 7:22
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    \$\begingroup\$ @jonk What a great source! \$\endgroup\$
    – Neil_UK
    Apr 8, 2020 at 8:30
  • \$\begingroup\$ @Neil_UK Yeah. I liked it, too. It's a nice segue into the ideas involved. \$\endgroup\$
    – jonk
    Apr 8, 2020 at 9:02
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    \$\begingroup\$ finding open-circuit current and short-circuit voltage - don't you mean the other way round? \$\endgroup\$
    – Andy aka
    Apr 8, 2020 at 9:14

1 Answer 1

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By opening and shorting the port, you are finding the extreme ends of the I_V curve.

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