Kirchhoffsche regeln (Parallel- und Reihenschaltung)?
Ich sollte für diese Aufgabe I_1 bis I_3 berechnen mithilfe der Knotenregel und Maschensatz (oben und unten).
Das Problem ist halt das meine Ergebnisse sehr seltsam sind. Ich habe schon viermal die Aufgabe durchgerechnet und bekomme immer wieder unterschiedliche Ergebnisse raus.
Die I’s die ich berechnet habe, sind viel zu groß (5 bis 14 Ampere) was eigentlich viel zu viel ist. Ich habe daraufhin mein Tutor gefragt, er meinte nur das es eigentlich nicht üer einen Ampere groß sein soll.
Kann mir wer helfen wie ihr den Knotensatz und Maschensatz anwenden würdet? Weil ich bin mir ziemlich sicher das ich falsch vorgehe, aber ich weiß halt nie was genau.
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I would not use the set of nodes or the set of stitches for calculation. (I never use the mesh set anyway.)
At first, I would summarize the resistors connected in series. Then there are three voltage sources with one resistance in series.
If one would now know the voltage between the left node (R3, R4) and the right node (R1, R2, R5), one could calculate the voltages at the three resistors and from them the currents.
In order to determine this voltage, the three voltage sources can be converted with their resistors into current sources. Then there are three parallel-connected current sources and three parallel-connected resistors. It can be summarized and the desired voltage can be calculated.
This allows the voltages at the three resistors to be calculated and the currents from this.
The node set is then useful as a sample.
Node set is simple – we have only 2 nodes here, which obviously each yield
-I1 + I2 + I3 = 0
In the case of the 3 stitches it is somewhat more confusing – one has to be careful that one does not get lost with the signs (one direction of the circulation around the mesh, on the other hand current directions and directions of the voltage sources). Here I’d check again if I got everything right.
For the lower mesh (I2 and I3) I would have (started with the left knot, counterclockwise)
I3 R4 – U3 + I3 r3 + I3 R5 – I2 R2 – I2 r2 + U2 = 0