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if circuit on the left of terminal A-B are in black box and I want to find Thevenin resister by apply voltage source v1 to the terminal A-B measure current i1 then change voltage source to v2 and measure current i2. But R=v1-v2/i1-i2 = infinity because i1=i2. How can I find Thevenin equivalent circuit?

  • \$\begingroup\$ But for this circuit the Rth is equal to infinity. \$\endgroup\$
    – G36
    Nov 22, 2020 at 11:53
  • \$\begingroup\$ is it correct to be infinity? if Rth is infinity so current should be zero \$\endgroup\$ Nov 22, 2020 at 11:57
  • \$\begingroup\$ Because you decided to use an ideal current source. So don't be surprised to encounter an paradox here. \$\endgroup\$
    – G36
    Nov 22, 2020 at 12:02
  • \$\begingroup\$ You're right in being puzzled. In fact Thevenin equivalent of an ideal current source doesn't exist at all. And dually, the Norton one of an ideal voltage source doesn't either. It's just this simple. \$\endgroup\$
    – carloc
    Nov 22, 2020 at 12:43

1 Answer 1


How can I find Thevenin equivalent circuit?

A resistor in series with a constant current source has no effect. The current source projects an impedance of infinite ohms and any resistor in series with it does not change infinity to anything other than er... infinity.

  • \$\begingroup\$ i do not understand.can you explain more? \$\endgroup\$ Nov 22, 2020 at 12:03
  • \$\begingroup\$ What specifically don't you understand? I could try and guess what you don't understand but I'm not going to do that because this is a question and answer site and not a talking shop or forum. \$\endgroup\$
    – Andy aka
    Nov 22, 2020 at 12:06
  • \$\begingroup\$ why does current source project an impedance of infinite ohms,because there is 1 ma current flow \$\endgroup\$ Nov 22, 2020 at 12:12
  • 1
    \$\begingroup\$ A current source of 1 mA could be said to be equivalent to 1 volt in series with 1 kohm. Into a short circuit is generates 1 mA but, into a load of 3 kohm (for example), the current is only 0.25 mA i.e. not a great current source. So, maybe we make a 1 mA current source from 100 volts in series with 100 kohm. It still produces 1 mA into a short but into a 3 kohm load it produces 0.97 mA. It's not too bad now but it's still not perfect so, maybe if we make the current source from 10,000 volt and 10 Mohm it's better; into a 3 kohm load the current is 0.9997 mA. Do you see where this is going? \$\endgroup\$
    – Andy aka
    Nov 22, 2020 at 12:23

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