Three Resistors
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:
Calculate the possible resistances you can get by combining three resistors with resistances RaO RbO and RcO using one two or all three.
Solution:
Most combinations are straightforward: itemize item One resistor: Ra Rb Rc item Two resistors in series: itemize itemast Ra+RbRasbP itemast Rb+RcRbscP itemast Rc+RaRcsaP itemize item Two resistors in parallel: itemize itemast left/Ra+/Rbright^-RapbP itemast left/Rb+/Rcright^-RbpcP itemast left/Rc+/Raright^-RcpaP itemize item Three resistors in series: Ra+Rb+RcRabcsP item Three resistors in parallel: left/Ra+/Rb+/Rcright^-RabcpP itemize There are six more combinations with three resistors where either two resistors in series are in parallel to the third resistor or two resistors in parallel are in series to the third resistor see figure. center includegraphicswidthtextwidth#image_path:threresistors# center The resistance for the circuits in the first row can be calculated as follows: sscRtot RAabpcF With the different values for R_ R_ and R_ we get RAabpcP RAbcpaP and RAcapbP respectively. For the circuits in the second row the resitance can be calculated as follows: sscRtot RBasbcF The resulting resistances are RBasbcP RBbscaP and RBcsabP respectively.
Calculate the possible resistances you can get by combining three resistors with resistances RaO RbO and RcO using one two or all three.
Solution:
Most combinations are straightforward: itemize item One resistor: Ra Rb Rc item Two resistors in series: itemize itemast Ra+RbRasbP itemast Rb+RcRbscP itemast Rc+RaRcsaP itemize item Two resistors in parallel: itemize itemast left/Ra+/Rbright^-RapbP itemast left/Rb+/Rcright^-RbpcP itemast left/Rc+/Raright^-RcpaP itemize item Three resistors in series: Ra+Rb+RcRabcsP item Three resistors in parallel: left/Ra+/Rb+/Rcright^-RabcpP itemize There are six more combinations with three resistors where either two resistors in series are in parallel to the third resistor or two resistors in parallel are in series to the third resistor see figure. center includegraphicswidthtextwidth#image_path:threresistors# center The resistance for the circuits in the first row can be calculated as follows: sscRtot RAabpcF With the different values for R_ R_ and R_ we get RAabpcP RAbcpaP and RAcapbP respectively. For the circuits in the second row the resitance can be calculated as follows: sscRtot RBasbcF The resulting resistances are RBasbcP RBbscaP and RBcsabP respectively.
Meta Information
Exercise:
Calculate the possible resistances you can get by combining three resistors with resistances RaO RbO and RcO using one two or all three.
Solution:
Most combinations are straightforward: itemize item One resistor: Ra Rb Rc item Two resistors in series: itemize itemast Ra+RbRasbP itemast Rb+RcRbscP itemast Rc+RaRcsaP itemize item Two resistors in parallel: itemize itemast left/Ra+/Rbright^-RapbP itemast left/Rb+/Rcright^-RbpcP itemast left/Rc+/Raright^-RcpaP itemize item Three resistors in series: Ra+Rb+RcRabcsP item Three resistors in parallel: left/Ra+/Rb+/Rcright^-RabcpP itemize There are six more combinations with three resistors where either two resistors in series are in parallel to the third resistor or two resistors in parallel are in series to the third resistor see figure. center includegraphicswidthtextwidth#image_path:threresistors# center The resistance for the circuits in the first row can be calculated as follows: sscRtot RAabpcF With the different values for R_ R_ and R_ we get RAabpcP RAbcpaP and RAcapbP respectively. For the circuits in the second row the resitance can be calculated as follows: sscRtot RBasbcF The resulting resistances are RBasbcP RBbscaP and RBcsabP respectively.
Calculate the possible resistances you can get by combining three resistors with resistances RaO RbO and RcO using one two or all three.
Solution:
Most combinations are straightforward: itemize item One resistor: Ra Rb Rc item Two resistors in series: itemize itemast Ra+RbRasbP itemast Rb+RcRbscP itemast Rc+RaRcsaP itemize item Two resistors in parallel: itemize itemast left/Ra+/Rbright^-RapbP itemast left/Rb+/Rcright^-RbpcP itemast left/Rc+/Raright^-RcpaP itemize item Three resistors in series: Ra+Rb+RcRabcsP item Three resistors in parallel: left/Ra+/Rb+/Rcright^-RabcpP itemize There are six more combinations with three resistors where either two resistors in series are in parallel to the third resistor or two resistors in parallel are in series to the third resistor see figure. center includegraphicswidthtextwidth#image_path:threresistors# center The resistance for the circuits in the first row can be calculated as follows: sscRtot RAabpcF With the different values for R_ R_ and R_ we get RAabpcP RAbcpaP and RAcapbP respectively. For the circuits in the second row the resitance can be calculated as follows: sscRtot RBasbcF The resulting resistances are RBasbcP RBbscaP and RBcsabP respectively.
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