1. Kirchhoffsche Regel/Knotenregel?
Guten Abend,
Ich bräuchte unbedingt Mal eure Hilfe da ich das einfach nicht in mein Kopf reinbekomme.
Wie oder was muss ich bei den folgenden Aufgaben machen?
A)Stellen sie die Knotenregel zur Berechnung der Ströme für die Schaltung nach BILD 1 (folgendes Bild) auf und
b) berechnen Sie daraus die Stromstärke i3.
Bin Grad total überfordert und frag mich ob ich jetzt I1 + i2 und ich dann i3 raus bekomme oder ob ich das ganze andersherum tuhen muss?!..
(No Ratings Yet)
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The knot rule says everything that goes inside must come out again.
Imagine the node as a bucket into which you pump water and out again.
All currents in which the arrows point pure into the nodes become water which is pumped in and all currents in which the arrows are pointing away from the node to water which is pumped out.
I1 and I2 go in, I4 and I5 out.
Question how much electricity is now “in the bucket”?
The olle Kirchhoff says at the end the bucket must be empty. So how much electricity do you have to put in the bucket with I3 so that it gets empty?
(not surprising, the model with the bucket and the water is pretty stupid, so you have to take it that you get negative numbers as an intermediate result, which is naturally not possible with water! But just keep counting bravely, in the end, then tuning again)
The egg was extremely helpful, so it left 0.6A at i3…
In the next task, however, I will be asked what I would conclude when -0,6A comes out. I’m following Dan that I did something wrong or not?
Nö, there you did nix wrong, the (actual) current just flowed against the 8-voluntary) arrow direction!
in the real world of an eclectic or electronicsman, by the way, this happens constantly, because for reasons of simplification, one does not make a head at all about which direction the current really flows. At the first node of the circuit you simply draw ALL current arrows in the direction of the node and slope when there are several nodes along the arrows from one node to the other. All currents additionally encountered at the ninth node are simply drawn again with the arrow to the node. And then continue to the next node until you’ve all been through. With this method you don’t have to think about it at all, but at the end you just keep reckoning. Okay, if you have a few negative values less, but the result won’t be more correct.
If the sign is contrary to the arrow at the result, then you conclude that the current flows in truth in the opposite direction and the direction of the arrow must be turned around in order to correctly represent the direction.
Salopp: You look at the point. All that points to this point, you count with Plus. All that goes away from the point, you count with minus. That is, you make the sum of all arrows and the Kirchhoff rule says that this sum is 0.
Now you have your knot equation. Because I3 is in demand, you simply convert it to I3, set the numerical values and enjoy your result.
I have actually done that:)
Rule: The sum of the currents to a node is 0:
1.9+1.1-2.7-0.9+i3=0
Solve the equation according to i3 and you get:
i3=0.6A
Thank you.