Kirchhoff I
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
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Video
\(\LaTeX\)
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Exercise:
Bestimmen Sie die Ströme in den verschiedenen Widerständen vgl. Abb.. center tikzpicturescale. draw thick -- -- -- -- -- .; draw thick - -- ; draw thick -.. -- ..; node at -.. U_ V; draw thick -- ; draw thick -- .; node at . U_ V; draw thick -- ; draw thick .. -- ..; draw thickfillwhite .. rectangle node rightxshift.cm R_ Omega ..; draw thick fillwhite . rectangle node aboveyshift .cm R_ Omega.; draw thick fillwhite . rectangle node aboveyshift .cm R_ Omega.; draw thick fillwhite -. rectangle node belowyshift -.cm R'_ Omega.; draw thick fillwhite -. rectangle node belowyshift -.cm R'_ Omega.; draw fillblack circle .mm node aboveyshiftmm a; draw fillblack circle .mm node belowyshift-mm b; draw very thick- -- node above I_ ; draw very thick- -- node above I_ ; draw very thick- . -- node right I_ ; tikzpicture center
Solution:
Mit der Knotenregel finden wir für den oberen Knoten I_ I_ + I_. Betrachten wir nun die linke Masche und durchlaufen wir sie im Uhrzeigersinn so erhalten wir: U_ I_R_ + U_ + I_R_ + I_R'_. Analog für die rechte Masche U_ I_R_ + I_R'_ - I_R_. Wir definieren R_ + R'_ R_' und R_ + R'_ R_' dann erhalten wir eqnarray* U_ & I_ R_' - I_R_ quadRightarrow quad I_ fracU_ + I_R_R_' U_ - U_ & I_ R_' + I_R_ quadRightarrowquad I_ fracU_ - U_ - I_R_R_' eqnarray* Eingesetzt in die erste Gleichung ergibt: eqnarray* fracU_ - U_ - I_R_R_' & fracU_ + I_R_R_' + I_mm R_'U_ - U_ - I_R_ & R_'U_ + I_R_ + I_R_'R_'mm I_ & fracU_-U_R_' - U_R_'R_'R_+R_'R_' +R_'R_ A. eqnarray* Damit bekommt man für I_ A und für I_ A.
Bestimmen Sie die Ströme in den verschiedenen Widerständen vgl. Abb.. center tikzpicturescale. draw thick -- -- -- -- -- .; draw thick - -- ; draw thick -.. -- ..; node at -.. U_ V; draw thick -- ; draw thick -- .; node at . U_ V; draw thick -- ; draw thick .. -- ..; draw thickfillwhite .. rectangle node rightxshift.cm R_ Omega ..; draw thick fillwhite . rectangle node aboveyshift .cm R_ Omega.; draw thick fillwhite . rectangle node aboveyshift .cm R_ Omega.; draw thick fillwhite -. rectangle node belowyshift -.cm R'_ Omega.; draw thick fillwhite -. rectangle node belowyshift -.cm R'_ Omega.; draw fillblack circle .mm node aboveyshiftmm a; draw fillblack circle .mm node belowyshift-mm b; draw very thick- -- node above I_ ; draw very thick- -- node above I_ ; draw very thick- . -- node right I_ ; tikzpicture center
Solution:
Mit der Knotenregel finden wir für den oberen Knoten I_ I_ + I_. Betrachten wir nun die linke Masche und durchlaufen wir sie im Uhrzeigersinn so erhalten wir: U_ I_R_ + U_ + I_R_ + I_R'_. Analog für die rechte Masche U_ I_R_ + I_R'_ - I_R_. Wir definieren R_ + R'_ R_' und R_ + R'_ R_' dann erhalten wir eqnarray* U_ & I_ R_' - I_R_ quadRightarrow quad I_ fracU_ + I_R_R_' U_ - U_ & I_ R_' + I_R_ quadRightarrowquad I_ fracU_ - U_ - I_R_R_' eqnarray* Eingesetzt in die erste Gleichung ergibt: eqnarray* fracU_ - U_ - I_R_R_' & fracU_ + I_R_R_' + I_mm R_'U_ - U_ - I_R_ & R_'U_ + I_R_ + I_R_'R_'mm I_ & fracU_-U_R_' - U_R_'R_'R_+R_'R_' +R_'R_ A. eqnarray* Damit bekommt man für I_ A und für I_ A.
Meta Information
Exercise:
Bestimmen Sie die Ströme in den verschiedenen Widerständen vgl. Abb.. center tikzpicturescale. draw thick -- -- -- -- -- .; draw thick - -- ; draw thick -.. -- ..; node at -.. U_ V; draw thick -- ; draw thick -- .; node at . U_ V; draw thick -- ; draw thick .. -- ..; draw thickfillwhite .. rectangle node rightxshift.cm R_ Omega ..; draw thick fillwhite . rectangle node aboveyshift .cm R_ Omega.; draw thick fillwhite . rectangle node aboveyshift .cm R_ Omega.; draw thick fillwhite -. rectangle node belowyshift -.cm R'_ Omega.; draw thick fillwhite -. rectangle node belowyshift -.cm R'_ Omega.; draw fillblack circle .mm node aboveyshiftmm a; draw fillblack circle .mm node belowyshift-mm b; draw very thick- -- node above I_ ; draw very thick- -- node above I_ ; draw very thick- . -- node right I_ ; tikzpicture center
Solution:
Mit der Knotenregel finden wir für den oberen Knoten I_ I_ + I_. Betrachten wir nun die linke Masche und durchlaufen wir sie im Uhrzeigersinn so erhalten wir: U_ I_R_ + U_ + I_R_ + I_R'_. Analog für die rechte Masche U_ I_R_ + I_R'_ - I_R_. Wir definieren R_ + R'_ R_' und R_ + R'_ R_' dann erhalten wir eqnarray* U_ & I_ R_' - I_R_ quadRightarrow quad I_ fracU_ + I_R_R_' U_ - U_ & I_ R_' + I_R_ quadRightarrowquad I_ fracU_ - U_ - I_R_R_' eqnarray* Eingesetzt in die erste Gleichung ergibt: eqnarray* fracU_ - U_ - I_R_R_' & fracU_ + I_R_R_' + I_mm R_'U_ - U_ - I_R_ & R_'U_ + I_R_ + I_R_'R_'mm I_ & fracU_-U_R_' - U_R_'R_'R_+R_'R_' +R_'R_ A. eqnarray* Damit bekommt man für I_ A und für I_ A.
Bestimmen Sie die Ströme in den verschiedenen Widerständen vgl. Abb.. center tikzpicturescale. draw thick -- -- -- -- -- .; draw thick - -- ; draw thick -.. -- ..; node at -.. U_ V; draw thick -- ; draw thick -- .; node at . U_ V; draw thick -- ; draw thick .. -- ..; draw thickfillwhite .. rectangle node rightxshift.cm R_ Omega ..; draw thick fillwhite . rectangle node aboveyshift .cm R_ Omega.; draw thick fillwhite . rectangle node aboveyshift .cm R_ Omega.; draw thick fillwhite -. rectangle node belowyshift -.cm R'_ Omega.; draw thick fillwhite -. rectangle node belowyshift -.cm R'_ Omega.; draw fillblack circle .mm node aboveyshiftmm a; draw fillblack circle .mm node belowyshift-mm b; draw very thick- -- node above I_ ; draw very thick- -- node above I_ ; draw very thick- . -- node right I_ ; tikzpicture center
Solution:
Mit der Knotenregel finden wir für den oberen Knoten I_ I_ + I_. Betrachten wir nun die linke Masche und durchlaufen wir sie im Uhrzeigersinn so erhalten wir: U_ I_R_ + U_ + I_R_ + I_R'_. Analog für die rechte Masche U_ I_R_ + I_R'_ - I_R_. Wir definieren R_ + R'_ R_' und R_ + R'_ R_' dann erhalten wir eqnarray* U_ & I_ R_' - I_R_ quadRightarrow quad I_ fracU_ + I_R_R_' U_ - U_ & I_ R_' + I_R_ quadRightarrowquad I_ fracU_ - U_ - I_R_R_' eqnarray* Eingesetzt in die erste Gleichung ergibt: eqnarray* fracU_ - U_ - I_R_R_' & fracU_ + I_R_R_' + I_mm R_'U_ - U_ - I_R_ & R_'U_ + I_R_ + I_R_'R_'mm I_ & fracU_-U_R_' - U_R_'R_'R_+R_'R_' +R_'R_ A. eqnarray* Damit bekommt man für I_ A und für I_ A.
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