Schaltung
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:
Betrachten Sie die folge Schaltung center circuitikzscale. european voltages draw to battery lvolt -- togeneric lOmega togeneric lOmega . . togeneric lOmega togeneric lOmega -- -- toshort iI_ . . toshort iI_ toshort iI_ ; circuitikz center enumerate item Bestimmen Sie den Ersatzwiderstand. item Bestimmen Sie die Stromstärken I_ I_ I_. item Bestimmen Sie die Spannung über dem Omega-Widerstand. enumerate
Solution:
enumerate item Ersatzwiderstand: Die Parallelschaltung ergibt frac+fracfrac to Omega Dazu addiert man den Omega Widerstand und erhält R_ersOmega item Es fliesst also U_R_ers I_ longrightarrow I_fracU_R_ers fracVOmega siA Das Verhältnis der Widerstände ist fracOmegaOmegafrac somit sind die Stromstärken im Verhältnis frac. Das geht gerade schön auf: I_fracsiA I_fracsiA item Die Spannung über dem Omega-Widerstand ergibt sich aus UR_ I_ Omega siA V enumerate Man hätte auch zuerst diese V herausfinden können und dann weiss man dass über dem parallelen Abschnitt die Spannung V herrscht V+VV und so hätte man auch die Ströme I_ I_ errechnen können.
Betrachten Sie die folge Schaltung center circuitikzscale. european voltages draw to battery lvolt -- togeneric lOmega togeneric lOmega . . togeneric lOmega togeneric lOmega -- -- toshort iI_ . . toshort iI_ toshort iI_ ; circuitikz center enumerate item Bestimmen Sie den Ersatzwiderstand. item Bestimmen Sie die Stromstärken I_ I_ I_. item Bestimmen Sie die Spannung über dem Omega-Widerstand. enumerate
Solution:
enumerate item Ersatzwiderstand: Die Parallelschaltung ergibt frac+fracfrac to Omega Dazu addiert man den Omega Widerstand und erhält R_ersOmega item Es fliesst also U_R_ers I_ longrightarrow I_fracU_R_ers fracVOmega siA Das Verhältnis der Widerstände ist fracOmegaOmegafrac somit sind die Stromstärken im Verhältnis frac. Das geht gerade schön auf: I_fracsiA I_fracsiA item Die Spannung über dem Omega-Widerstand ergibt sich aus UR_ I_ Omega siA V enumerate Man hätte auch zuerst diese V herausfinden können und dann weiss man dass über dem parallelen Abschnitt die Spannung V herrscht V+VV und so hätte man auch die Ströme I_ I_ errechnen können.
Meta Information
Exercise:
Betrachten Sie die folge Schaltung center circuitikzscale. european voltages draw to battery lvolt -- togeneric lOmega togeneric lOmega . . togeneric lOmega togeneric lOmega -- -- toshort iI_ . . toshort iI_ toshort iI_ ; circuitikz center enumerate item Bestimmen Sie den Ersatzwiderstand. item Bestimmen Sie die Stromstärken I_ I_ I_. item Bestimmen Sie die Spannung über dem Omega-Widerstand. enumerate
Solution:
enumerate item Ersatzwiderstand: Die Parallelschaltung ergibt frac+fracfrac to Omega Dazu addiert man den Omega Widerstand und erhält R_ersOmega item Es fliesst also U_R_ers I_ longrightarrow I_fracU_R_ers fracVOmega siA Das Verhältnis der Widerstände ist fracOmegaOmegafrac somit sind die Stromstärken im Verhältnis frac. Das geht gerade schön auf: I_fracsiA I_fracsiA item Die Spannung über dem Omega-Widerstand ergibt sich aus UR_ I_ Omega siA V enumerate Man hätte auch zuerst diese V herausfinden können und dann weiss man dass über dem parallelen Abschnitt die Spannung V herrscht V+VV und so hätte man auch die Ströme I_ I_ errechnen können.
Betrachten Sie die folge Schaltung center circuitikzscale. european voltages draw to battery lvolt -- togeneric lOmega togeneric lOmega . . togeneric lOmega togeneric lOmega -- -- toshort iI_ . . toshort iI_ toshort iI_ ; circuitikz center enumerate item Bestimmen Sie den Ersatzwiderstand. item Bestimmen Sie die Stromstärken I_ I_ I_. item Bestimmen Sie die Spannung über dem Omega-Widerstand. enumerate
Solution:
enumerate item Ersatzwiderstand: Die Parallelschaltung ergibt frac+fracfrac to Omega Dazu addiert man den Omega Widerstand und erhält R_ersOmega item Es fliesst also U_R_ers I_ longrightarrow I_fracU_R_ers fracVOmega siA Das Verhältnis der Widerstände ist fracOmegaOmegafrac somit sind die Stromstärken im Verhältnis frac. Das geht gerade schön auf: I_fracsiA I_fracsiA item Die Spannung über dem Omega-Widerstand ergibt sich aus UR_ I_ Omega siA V enumerate Man hätte auch zuerst diese V herausfinden können und dann weiss man dass über dem parallelen Abschnitt die Spannung V herrscht V+VV und so hätte man auch die Ströme I_ I_ errechnen können.
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