(50 points reward) according to Thevenin's theorem, how does the parallel connection of current source + resistance and voltage source + resistance become the form of voltage resistance?

(50 points reward) according to Thevenin's theorem, how does the parallel connection of current source + resistance and voltage source + resistance become the form of voltage resistance?

First determine the open circuit voltage VOC, and then determine R. there are three methods to determine R: directly through the open circuit to calculate R, for simple circuits; using the open short circuit method to calculate the short circuit current ISC, r = VOC / ISC; finally, the external voltage source or current source to calculate the V / I relationship. This is OK

In the basic circuit diagram, | --- means that the connecting wire is not disconnected, [] means a resistance of 1000 ohm. This problem is solved by using Thevenin's theorem
The correct answer is that the voltage of AB terminal is 25V
+30V
|
|---[]----|
[] |
|---[]----|---¤a
|
+
20V
-
|------------¤b
|
-Indicates grounding
The equivalent resistance is 500 ohm. The picture above is rough. You can see the illustration!

When you ask questions on your mobile phone, you can only input 99 words, not illustrations
Main circuit 15mA, branch current 5mA
Then VAB = 1K * 5mA + 20 = 25V
Or VAB = 30-1k * 5mA = 25V
Equivalent resistance req = two 1K in parallel = 500 ohm
I wrote the answer on paper, took a good picture and wanted to upload it. As a result, you asked with your mobile phone

When the resistance of a circuit is increased by 15 Ω and the current is 1 / 4 of the original, what is the original resistance?
If the current is to be 1 / 8 of its original value, how many ohm will be added?

Since the voltage of the circuit is constant, u = IR must be constant
Then there is: IR = 1 / 4I * (R + 15), the solution is: r = 5, that is, the original resistance is 5 Ω
If the current becomes 1 / 8, then the equation:
I * 5 = 1 / 8i * (5 + x) can be solved as follows: x = 35, that is to say, 35 Ω will be added~

When a 3 ohm resistor is connected in series in the circuit, the current becomes 5 / 6 of the original current, and the current becomes 1 / 2 of the original current in the circuit
When a 3-ohm resistor is connected in series in a circuit, if the current becomes 5 / 6 of the original resistance, how many ohm is the original resistance in the circuit? If the current becomes 1 / 2 of the original resistance, how many ohm should be connected in series in the circuit

When the power supply voltage of the circuit is 15V, the original resistance is 15 Ω and the current is 1a. When a 3-ohm resistance is connected in series, the current becomes 5 / 6 of the original resistance, the total resistance is 18 Ω and the current is 0.83333a; then the original resistance in the circuit is replaced by 27 Ω, that is, 27 + 3 = 30 (Ω), and the current becomes 1 / 2 of the original 1a, that is, 0.5A; then the total resistance in series is 30 Ω