Can someone calculate all the partial currents and partial voltages for me?
20 V. r total is 18.3
20 V. r total is 18.3
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Here’s the solution on the silver platter. But I just did it because it’s my hobby. The other commentators are right. You have to learn to do this yourself.
So: Starting from the redrawn circuit, you define all stitches (red) and nodes (blue).
and formulates all 10 stitch equations (voltage equations) and all 7 node equations (current equations).
You see that not all 20 currents occur in the equations. In series circuits it is known that the currents are always the same with resistors connected in series. Furthermore, in another task, you have been able to see that all resistors have a uniform value of 10 ohms. Therefore, you can also make current equations from the stitch equations via the Ohmic Law. The voltage drop in [V] now has the tenfold numerical value of the current measured in [A]. So you come to the following equation system.
Here you can immediately shorten all the factors 10. Furthermore, the currents i1, i6, i11 and i20 can be expressed by other identical currents. In this way you can reduce the number of variables. The preparation results in the following equation system consisting of 17 equations with 16 unknowns, the total current i_G also being determined. An equation will prove superfluous.
For the resolution of the equation system, the system is represented in matrix form.
The triangulation of the equation system thus shown leads to the following matrix representation.
This system allows the convenient calculation of all currents, starting with the total current i_G = 1,092A. Thus, the overall resistance is also
confirmed.
The remaining currents can be calculated step-up backwards. The associated voltages are ten times measured in [volt] and are well represented in the following drawing.
“r total is 18.3”.
This is a very interesting result. Firstly, the unit (Ohm, kOhm, mOhm,…) and secondly surprises the indication of a numerical value, although there are no resistance values.
Well, there were already some people helpful (incl. me) but somehow it’s enough. Not every student should be maneuvered by third parties while maintaining total unawareness, in order to lose burden on deceased employers.
Yeah, I’m sure someone can do that. But maybe you should start. On GF there are many helpful people, but few who like to do tasks for people who are lazy.