Vth voltage in thevenin theorem

[PG1995]

Hi :smile:

Please have a look on Example 4.8. There are two scans.

1: https://img585.imageshack.us/img585/6673/understandingthevenin1.jpg
2: https://img818.imageshack.us/img818/3301/understandingthevenin2.jpg

First we find the equivalent resistance of the circuit on the left of terminals "a" and "b". The equivalent resistance comes to be 4Ω. In other words, we can replace the circuit part lying on left of terminals "a" and "b" with only one resistor of 4Ω. I hope I have it right.

Then we find Vth voltage which is appearing around terminals "a" and "b". This is the voltage which can be detected by a voltmeter around the terminals a-b. Vth is 30V.

Now have a look on Figure 4.29 in scan #2. We have 4Ω resistor (Rth, or equivalent resistance) in series with 30V DC source and RL (load resistor) which is connected to terminals a-b. It is obvious that some of the Vth, 30V, is going to drop around 4Ω resistor which means now the voltage which will appear around the terminals "a" and "b" will be less than 30V. Originally we found the Vth as the voltage which appears around the terminals a-b but now voltage which is appearing on terminals a-b is not Vth. Why is so? I hope you understand my question. Could you please tell me? Thanks.

 

[KerimF]

Good question since its answer will tell us why we bother ourselves to find Rth and Vth in the first place.

On the figure 4.27, the circuit of interest has 5 items (excluding RL which is considered as an external one).

Let us suppose we skipped the calculations of Rth and Vth, and we like to find the current through RL=6Ω. The circuit would consist of 3 loops (I1, I2, and I3 starting from the left) so we will have 3 equations and the magnitude of the load current will be the value of I3 (Note: if I3 happens to have the same direction of the requested current I then I=I3 otherwise I=-I3). If you will do this exercise you will end up with I=3A.

If after this lengthy work, one may ask... what could be the current of the load if RL=16Ω ? To answer him, we need to repeat ALL the previous steps (that is solving the equations of 3 loops) to find out, at last, that the answer is I=1.5A.

So the purpose in finding an EQUIVALENT circuit of a rather complex one is to do the hard work just once so that we don't need to repeat it every time the external load is changed.

One of the methods to find an equivalent circuit is based on Thevenin theorem which helps us replace ALL the branches and loops of a circuit (seen from two terminals, as a and b) by a voltage source (may be called Vth) and a resistance in series (may be called Rth).

As you see, after we worked hard to find Vth and Rth, we can forget the rather complex circuit we start with (at the left of a and b) and replace it by Vth and Rth only (how nice!). Now with RL, we have only one loop that consists of Vth, Rth and RL. So changing the value of RL needs us just to repeat the calculation of a single loop which gives I = 30 / (4 + RL) in our case. And we just replace RL by its value to get the answer (we don't need to solve the original 3 loops).

Imagine if the original circuit has 50 or 100 loops. Isn't very nice to replace them by just Vth and Rth? And when we know the value of the load resistance, it is a matter of seconds to get the output voltage and current (Note: In real DC circuits the output voltage is always lower than Vth because of the voltage drop on Rth, if Rth is small the difference will be relatively small too and zero if the load current is zero that is open circuit).

Kerim