Northon and Thevenin theorems

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Thevenin theorem

I will explain it with an example. Look at a ciruit at picture 1. E is the voltage source, and R_int is its internal resistance. We need to find the current I_L, which passes through the resistance R.

Lets make the following mental operation: remove R from the circuit (pic. 2). We need to find the voltage between points A and B (U_ab). The resistance of the circuit through which current I passes, is equal to R_int+R₁+ R₂

Therefore, the current I will be equal (pic. 3, 1). According to Ohm's law the voltage U_ab will be equal to(see pic. 3, 2).

We now know the voltage U_ab and the resistance R. Maybe we can calculate the current I_L in accordance with Ohms law?

Now let us mentally replace the voltage source E with its internal resistance R_int (pic. 5, 1 ). The resistance between points A and B (we denote it as R_t) is the resistance of two parallel-connected branches: R_int+R₁ and R₂. It will be equal to (see pic. 5, 2)

For the solution of our problem (calculate the current I_L), we can consider the circuits in figure (6, 1) and (6, 2) as equivalent. This is the essence of the Thevenin theorem. That is, the entire circuit (6, 1) can be regarded as one battery with EMF U_ab and internal resistance R_t. Now we can easily calculate the current I_L (pic. 7, 2). The circuit in (pic. 6, 1) is very simple. But if was much more complicated, with lots of resistors and voltage sources , it could be considered as the equivalent to circuit (6, 2).

Northon theorem

Again I explain it with an example. Look at a circuit in picture 8. Our task is to calculate current I_L.

Like in Thevenin theorem, we will mentally remove resistance R. But now we will short-circuit points A and B, so that resistance between them is null (picture 9.). Let us calculate current I_n. To achieve that we will calculate current I_e first. In our scheme R₂ and R₃ are connected in parallel. Let us calculate their resistance. Then in accordance with Ohm law we find current I_e (picture. 9, 3). Then we calculate current I_n (picture 9, 4).

Now let us remove the jumper between A and B, and also replace voltage source E with a jumper. The internal resistance R_int will stay in place. (picture. 10, 1). We did something like this, when we learned the Thevenin theorem. Let us find resistance between points A and B. In our scheme R_int+R₁ and R₂ are connected in parallel. Let us find their resistance (picture 10, 2). The resistance R_n will be (see picture 10, 3).

The essence of the Northon theorem is that we can replace the initial circuit (picture 11, 1) with a much simpler one (picture 11, 2). It consists, apart from the load, the current through which we want to calculate, of a current source (which produces a constant current that we have calculated) and a resistance connected in parallel to it (which we also have calculated). If the initial circuit was more complicated, with many resistors and voltage sources, it could also be replaced with scheme (11, 2)

Let us calculate I_L now. The total resistance of the circuit (11, 2) will be (picture 12, 1). Hence the voltage between points A and B will be (12, 2). Current I_L will be (pic. 12, 3).