By using Thevenin's theorem and Norton's theorem, the graphic circuit is transformed into an equivalent voltage source and an equivalent current source, and the solution of open circuit voltage uo and short circuit current is explained I understand,

First, the power supply is equivalent, 5 Ω plus 5V on the left side can form a current source, 1A direction is positive (in the power supply, the current flows from low potential to high potential, so it is positive 2A - 1A = 1a, the current source is in parallel with 5 Ω resistance, and further it can be equivalent to a positive 5V voltage source and 5 Ω resistance in series

Davining's theorem of University circuit. Short circuit current method for calculating equivalent resistance

How to get the accurate open circuit voltage when applying the equivalent transformation of Thevenin's theorem

In general, it is assumed that there is an ideal voltmeter (infinite resistance) at the open circuit. In this way, the reading of the voltmeter is what you want!

On Thevenin's theorem and Norton's theorem
In the circuit solution, when Thevenin's theorem is used for equivalence, it is found that the internal resistance is 0. If the same circuit is solved by Norton's theorem, and the resistance is 0 and the current source is in parallel, it will be short circuited. What kind of circuit can't be applied to the two theorems at the same time? Thank you

The circuit equivalent to ideal voltage source can not be expressed by Norton's theorem,
The circuit equivalent to ideal current source cannot be expressed by Thevenin's theorem

By using Thevenin's theorem and Norton's theorem, the graphic circuit is transformed into an equivalent voltage source and an equivalent current source, and the solution of open circuit voltage uo and short circuit current is explained I understand,