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Ersatzwiderstand mit Symmetrie

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Exercise

https://texercises.com/exercise/ersatzwiderstand-mit-symmetrie/

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The following quantities appear in the problem: elektrischer Widerstand \(R\) / Federkonstante \(D\) /

The following formulas must be used to solve the exercise: \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \quad \) \(\frac{1}{D} = \frac{1}{D_1} + \frac{1}{D_2} \quad \) \(R = R_1 + R_2 \quad \)

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Exercise:
In der unten abgebildeten Schaltung sei R_R_pqOmega R_R_pqOmega. Zwischen den Punkten a und b ist eine Spannung von UpqV angelegt. Berechne den Strom I der von a nach b fliesst indem du zuerst den Ersatzwiderstand der Schaltung findest. center tikzpicture %node a at . draw rotate hspacemmRhspacemm ; drawthick --; drawthick --; drawthick --; drawthick --; drawthick ------; drawthick ------; drawthick ----; drawthick ----; nodeabove at a; nodeabove at b; drawthick -. rectangle .; drawthick -. rectangle .; drawthick -. rectangle .; drawthick . rectangle .; drawthick -. rectangle -.; node at R_; node at R_; node at R_; node at - R_; node at R_; tikzpicture center

Solution:
Aufgrund von Symmetrieüberlegungen wird klar dass durch den Widerstand R_ kein Strom fliesst. Er kann daher ausser acht gelassen werden. Die Schaltung besteht dann noch aus zwei Serieschaltungen von je einem pqOmega und einem pqOmega Widerstand die ihrerseits wieder parallel geschaltet sind. Insgesamt also R_ R_ + R_ pqOmega quadbzw. R_ R_ + R_ pqOmega. Eine Parallelschaltung aus zwei Widerständen von je pqOmega hat einen Widerstand von R leftfracR_+fracR_right^- pq.Omega. Der Gesamtstrom von a nach b ist damit nach dem Ohm'schen Gesetz I fracUR fracpqVpq.Omega pqA.

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Contained in these collections:

Ersatzwiderstand von Schaltung by TeXercises

8 | 15
Ersatzwiderstand 2 by uz

2 | 6

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Tags	ersatzwiderstand, gleichstromlehre, physik, symmetrie
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Difficulty	(3, default)
Points	2 (default)
Language	(Deutsch)
Type	Calculative / Quantity
Creator	uz
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