Advanced RL Circuit
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
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Exercise:
The circuit in the circuit diagram below consists of three resistors R_RaO R_RbO R_RcO and an inductor LLO connected to a VO voltage supply. center includegraphicswidth.mm#image_path:lr-circuit# center abcliste abc What are the currents through the three resistors I_ I_ and I_ and the voltage V_L across the inductor immediately after closing the switch? abc What are the currents and the voltage V_L after a very long time? abc What are the currents and the voltage V_L immediately after reopening the switch? abcliste
Solution:
abcliste abc Immediately after closing the switch there is no current through the inductor Lenz's law i.e. resultI_. It follows that I_I_. For the left loop containing the voltage supply and the resistors R_ and R_ we find V_ R_ I_+R_ I_ R_+R_ I_ I_ I_ IaF fracVRa+Rc resultIaP The voltage across R_ and L in series is the same as the voltage across R_ we find for the induced voltage across the inductor V_L R_ I_ - R_ I_ R_ I_-R_ VLF RctimesIa resultVLP abc After a long enough time the system has reached a steady state i.e. the current does not change any more. This means that the voltage across the inductor has disappeared: resultV_L. The circuit can be considered to be an ordinary resistor circuit and the inductur can be ignored. Resistors R_ and R_ are in parallel to each other and together in series to R_. The total resistance of the circuit is sscRtot R_+leftfracR_+fracR_right^- R_+fracR_ R_R_+R_ RtF The total current I_ is then I_ fracV_sscRtot IafF fracVtimesRb+RcRatimesRb+RatimesRc+RbtimesRc resultIafP The total current is split up between R_ and R_ in the inverse ratio of the resistances: I_ IbfF IaftimesfracRcRb+Rc resultIbfP I_ IcfF IaftimesfracRbRb+Rc resultIcfP abc Immediately after reopening the switch the current through the inductor is still equal to the steady-state current I_IbfP. The right loop is still closed but the left loop is open so we must have I_ result I_ I_ resultIbfP The induced voltage across the inductor is V_L R_ I_+R_ I_ VLroF Rb+RctimesIbfP resultVLroP abcliste
The circuit in the circuit diagram below consists of three resistors R_RaO R_RbO R_RcO and an inductor LLO connected to a VO voltage supply. center includegraphicswidth.mm#image_path:lr-circuit# center abcliste abc What are the currents through the three resistors I_ I_ and I_ and the voltage V_L across the inductor immediately after closing the switch? abc What are the currents and the voltage V_L after a very long time? abc What are the currents and the voltage V_L immediately after reopening the switch? abcliste
Solution:
abcliste abc Immediately after closing the switch there is no current through the inductor Lenz's law i.e. resultI_. It follows that I_I_. For the left loop containing the voltage supply and the resistors R_ and R_ we find V_ R_ I_+R_ I_ R_+R_ I_ I_ I_ IaF fracVRa+Rc resultIaP The voltage across R_ and L in series is the same as the voltage across R_ we find for the induced voltage across the inductor V_L R_ I_ - R_ I_ R_ I_-R_ VLF RctimesIa resultVLP abc After a long enough time the system has reached a steady state i.e. the current does not change any more. This means that the voltage across the inductor has disappeared: resultV_L. The circuit can be considered to be an ordinary resistor circuit and the inductur can be ignored. Resistors R_ and R_ are in parallel to each other and together in series to R_. The total resistance of the circuit is sscRtot R_+leftfracR_+fracR_right^- R_+fracR_ R_R_+R_ RtF The total current I_ is then I_ fracV_sscRtot IafF fracVtimesRb+RcRatimesRb+RatimesRc+RbtimesRc resultIafP The total current is split up between R_ and R_ in the inverse ratio of the resistances: I_ IbfF IaftimesfracRcRb+Rc resultIbfP I_ IcfF IaftimesfracRbRb+Rc resultIcfP abc Immediately after reopening the switch the current through the inductor is still equal to the steady-state current I_IbfP. The right loop is still closed but the left loop is open so we must have I_ result I_ I_ resultIbfP The induced voltage across the inductor is V_L R_ I_+R_ I_ VLroF Rb+RctimesIbfP resultVLroP abcliste
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Exercise:
The circuit in the circuit diagram below consists of three resistors R_RaO R_RbO R_RcO and an inductor LLO connected to a VO voltage supply. center includegraphicswidth.mm#image_path:lr-circuit# center abcliste abc What are the currents through the three resistors I_ I_ and I_ and the voltage V_L across the inductor immediately after closing the switch? abc What are the currents and the voltage V_L after a very long time? abc What are the currents and the voltage V_L immediately after reopening the switch? abcliste
Solution:
abcliste abc Immediately after closing the switch there is no current through the inductor Lenz's law i.e. resultI_. It follows that I_I_. For the left loop containing the voltage supply and the resistors R_ and R_ we find V_ R_ I_+R_ I_ R_+R_ I_ I_ I_ IaF fracVRa+Rc resultIaP The voltage across R_ and L in series is the same as the voltage across R_ we find for the induced voltage across the inductor V_L R_ I_ - R_ I_ R_ I_-R_ VLF RctimesIa resultVLP abc After a long enough time the system has reached a steady state i.e. the current does not change any more. This means that the voltage across the inductor has disappeared: resultV_L. The circuit can be considered to be an ordinary resistor circuit and the inductur can be ignored. Resistors R_ and R_ are in parallel to each other and together in series to R_. The total resistance of the circuit is sscRtot R_+leftfracR_+fracR_right^- R_+fracR_ R_R_+R_ RtF The total current I_ is then I_ fracV_sscRtot IafF fracVtimesRb+RcRatimesRb+RatimesRc+RbtimesRc resultIafP The total current is split up between R_ and R_ in the inverse ratio of the resistances: I_ IbfF IaftimesfracRcRb+Rc resultIbfP I_ IcfF IaftimesfracRbRb+Rc resultIcfP abc Immediately after reopening the switch the current through the inductor is still equal to the steady-state current I_IbfP. The right loop is still closed but the left loop is open so we must have I_ result I_ I_ resultIbfP The induced voltage across the inductor is V_L R_ I_+R_ I_ VLroF Rb+RctimesIbfP resultVLroP abcliste
The circuit in the circuit diagram below consists of three resistors R_RaO R_RbO R_RcO and an inductor LLO connected to a VO voltage supply. center includegraphicswidth.mm#image_path:lr-circuit# center abcliste abc What are the currents through the three resistors I_ I_ and I_ and the voltage V_L across the inductor immediately after closing the switch? abc What are the currents and the voltage V_L after a very long time? abc What are the currents and the voltage V_L immediately after reopening the switch? abcliste
Solution:
abcliste abc Immediately after closing the switch there is no current through the inductor Lenz's law i.e. resultI_. It follows that I_I_. For the left loop containing the voltage supply and the resistors R_ and R_ we find V_ R_ I_+R_ I_ R_+R_ I_ I_ I_ IaF fracVRa+Rc resultIaP The voltage across R_ and L in series is the same as the voltage across R_ we find for the induced voltage across the inductor V_L R_ I_ - R_ I_ R_ I_-R_ VLF RctimesIa resultVLP abc After a long enough time the system has reached a steady state i.e. the current does not change any more. This means that the voltage across the inductor has disappeared: resultV_L. The circuit can be considered to be an ordinary resistor circuit and the inductur can be ignored. Resistors R_ and R_ are in parallel to each other and together in series to R_. The total resistance of the circuit is sscRtot R_+leftfracR_+fracR_right^- R_+fracR_ R_R_+R_ RtF The total current I_ is then I_ fracV_sscRtot IafF fracVtimesRb+RcRatimesRb+RatimesRc+RbtimesRc resultIafP The total current is split up between R_ and R_ in the inverse ratio of the resistances: I_ IbfF IaftimesfracRcRb+Rc resultIbfP I_ IcfF IaftimesfracRbRb+Rc resultIcfP abc Immediately after reopening the switch the current through the inductor is still equal to the steady-state current I_IbfP. The right loop is still closed but the left loop is open so we must have I_ result I_ I_ resultIbfP The induced voltage across the inductor is V_L R_ I_+R_ I_ VLroF Rb+RctimesIbfP resultVLroP abcliste
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