Kirchhoff mit Kapazität
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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Exercise:
Bestimmen Sie für das folge Schaltschema die Ladung Q auf der Kapazität C. center tikzpicturecircuit ee IEC %% Schaltung % Weg oben draw fillblack circle .mm -- to resistorinfoR_ -- ; draw fillblack -- - to batteryinfoU_ - to current direction'info'I_ - circle .mm; % Weg unten draw to batteryinfo'U_ - to resistorinfoR_ - to current direction'infoI_ -; % Weg diagonal draw to resistorinfoR_ .-. to current direction'info'I_ .-. to capacitorinfo'C .-. -- -; tikzpicture center Zahlenwerte: U_ V U_ V R_ ohm R_ ohm R_ ohm und C .mufarad
Solution:
Da auf der Diagonale eine Kapazität zwischen geschaltet ist fliesst durch diese Verbindung kein Strom d.h. I_ . Darauf folgt sofort I_ - I_. Für die äussere Masche im Uhrzeigersinn gilt dann: U_ - I_R_ - U_ - I_R_ . Aufgelöst nach I_ ergibt das: I_ fracU_-U_R_+R_ apx A. Damit ist I_ -A. Für die Ladung auf der Kapazität müssen wir zuerst die Spannung U_C über der Kapazität bestimmen. Für die linke untere Masche gilt im Uhrzeigersinn: U_ - U_C - I_R_ myRarrow U_C U_ - I_R_ apx V. Mit der Definition der Kapazität: C fracQU_C erhalten wir: Q C U_C apx muC.
Bestimmen Sie für das folge Schaltschema die Ladung Q auf der Kapazität C. center tikzpicturecircuit ee IEC %% Schaltung % Weg oben draw fillblack circle .mm -- to resistorinfoR_ -- ; draw fillblack -- - to batteryinfoU_ - to current direction'info'I_ - circle .mm; % Weg unten draw to batteryinfo'U_ - to resistorinfoR_ - to current direction'infoI_ -; % Weg diagonal draw to resistorinfoR_ .-. to current direction'info'I_ .-. to capacitorinfo'C .-. -- -; tikzpicture center Zahlenwerte: U_ V U_ V R_ ohm R_ ohm R_ ohm und C .mufarad
Solution:
Da auf der Diagonale eine Kapazität zwischen geschaltet ist fliesst durch diese Verbindung kein Strom d.h. I_ . Darauf folgt sofort I_ - I_. Für die äussere Masche im Uhrzeigersinn gilt dann: U_ - I_R_ - U_ - I_R_ . Aufgelöst nach I_ ergibt das: I_ fracU_-U_R_+R_ apx A. Damit ist I_ -A. Für die Ladung auf der Kapazität müssen wir zuerst die Spannung U_C über der Kapazität bestimmen. Für die linke untere Masche gilt im Uhrzeigersinn: U_ - U_C - I_R_ myRarrow U_C U_ - I_R_ apx V. Mit der Definition der Kapazität: C fracQU_C erhalten wir: Q C U_C apx muC.
Meta Information
Exercise:
Bestimmen Sie für das folge Schaltschema die Ladung Q auf der Kapazität C. center tikzpicturecircuit ee IEC %% Schaltung % Weg oben draw fillblack circle .mm -- to resistorinfoR_ -- ; draw fillblack -- - to batteryinfoU_ - to current direction'info'I_ - circle .mm; % Weg unten draw to batteryinfo'U_ - to resistorinfoR_ - to current direction'infoI_ -; % Weg diagonal draw to resistorinfoR_ .-. to current direction'info'I_ .-. to capacitorinfo'C .-. -- -; tikzpicture center Zahlenwerte: U_ V U_ V R_ ohm R_ ohm R_ ohm und C .mufarad
Solution:
Da auf der Diagonale eine Kapazität zwischen geschaltet ist fliesst durch diese Verbindung kein Strom d.h. I_ . Darauf folgt sofort I_ - I_. Für die äussere Masche im Uhrzeigersinn gilt dann: U_ - I_R_ - U_ - I_R_ . Aufgelöst nach I_ ergibt das: I_ fracU_-U_R_+R_ apx A. Damit ist I_ -A. Für die Ladung auf der Kapazität müssen wir zuerst die Spannung U_C über der Kapazität bestimmen. Für die linke untere Masche gilt im Uhrzeigersinn: U_ - U_C - I_R_ myRarrow U_C U_ - I_R_ apx V. Mit der Definition der Kapazität: C fracQU_C erhalten wir: Q C U_C apx muC.
Bestimmen Sie für das folge Schaltschema die Ladung Q auf der Kapazität C. center tikzpicturecircuit ee IEC %% Schaltung % Weg oben draw fillblack circle .mm -- to resistorinfoR_ -- ; draw fillblack -- - to batteryinfoU_ - to current direction'info'I_ - circle .mm; % Weg unten draw to batteryinfo'U_ - to resistorinfoR_ - to current direction'infoI_ -; % Weg diagonal draw to resistorinfoR_ .-. to current direction'info'I_ .-. to capacitorinfo'C .-. -- -; tikzpicture center Zahlenwerte: U_ V U_ V R_ ohm R_ ohm R_ ohm und C .mufarad
Solution:
Da auf der Diagonale eine Kapazität zwischen geschaltet ist fliesst durch diese Verbindung kein Strom d.h. I_ . Darauf folgt sofort I_ - I_. Für die äussere Masche im Uhrzeigersinn gilt dann: U_ - I_R_ - U_ - I_R_ . Aufgelöst nach I_ ergibt das: I_ fracU_-U_R_+R_ apx A. Damit ist I_ -A. Für die Ladung auf der Kapazität müssen wir zuerst die Spannung U_C über der Kapazität bestimmen. Für die linke untere Masche gilt im Uhrzeigersinn: U_ - U_C - I_R_ myRarrow U_C U_ - I_R_ apx V. Mit der Definition der Kapazität: C fracQU_C erhalten wir: Q C U_C apx muC.
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