Circuit analysis I. Sencillito.

Following a question in a forum it occurred to me to do some circuit simulation to exemplify some practical cases that are usually given. And how tension should be measured, for example. The specific case of this first example is as follows:

When measuring voltage at the terminals of a turbine it gives 0 v. If you disconnect the turbine it gives 220 v. You really can't know much with this little data. But if there is an important issue that we should all be clear about. To simplify I have used direct current and resistors instead of motors. It is not the same but it will comply perfectly to understand the concept. The simulations have been done with the Qucs program, which is under the GPL license.

We start from the following scheme:

R3 is the load (the turbine). A resistor does not behave the same as an engine and therefore the simulation values are not real. But it is good for what I want to demonstrate. The contact is open and there is no tension in R3 (Tensión2_2). (The program gives 2.2 * 10 evaded to -10, which is almost the same as saying 0.)

In this figure we can see the status when the contact is closed. The voltage in the contact (Tensión1_1) = 0 v. There is 2.2 Amp and the load reaches its 220 v.

In this case I have replaced the contact with a resistor to simulate that it is slightly damaged. There is a drop in tension in the contact, which we can measure perfectly and which is also generating heat.

Now the contact is very damaged, hardly any tension reaches the load (24.4 v).

All this roll was to get to this picture. We disconnect our turbine, the voltage drop in the contact disappears and all the voltage reaches the terminals of the turbine.

I know that for many this is a no-brainer but good. I think it may be interesting for some. To say that although it is made with the simulator, this is a simple scheme of a voltage divider, and the calculations can be perfectly performed with Ohmn's law and the sum of resistance in series.

FG_AUTHORS: David

See original.