LCR Parallel 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
Solution
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Exercise:
Show that the impedance and phase shift of an ac circuit with a resistor resistance R a capacitor capacitance C and a coil inductance L in parallel are given by * fracZ sqrtfracR^+leftomega C-fracomega Lright^ and * tanDeltaphi Rleftfracomega L-omega Cright H: Since the voltage is the same across all three elements in the parallel circuit you can start with the corresponding phasor in a phasor diagram and add the current phasors with their respective phaseshift.
Solution:
In the phasor diagram the current phasor for the resistor is in phase with the voltage while the current phasors for the capacitor and the coil have a phase shift of mp fracpi respectively see figure. center includegraphicswidth.mm#image_path:phasors-parallel-circuit# center The phasor for the total current I can be found by adding the phasors for the partial currents I_R I_C and I_L as vectors. The amplitude of the total current is I sqrtI_R^+leftI_C-I_Lright^ Using the relations I_R fracVR I_C fracVX_C I_L fracVX_L it follows for the impedance fracZ fracIV fracsqrtleftfracVRright^+leftfracVX_C-fracVX_Lright^V sqrtfracR^+leftfracX_C-fracX_Lright^ With the expressions for the reactances X_C and X_L we find fracZomega sqrtfracR^+leftomega C-fracomega Lright^ The phase shift is given by see figure tanDeltaphi fracI_L-I_CI_R fracfracVX_L-fracVX_CfracVR RleftfracX_L-fracX_Cright Rleftfracomega L-omega Cright Alternatively the expressions can be derived using the complex reactances tildeX_L and tildeX_C. For a parallel circuit the reciprocal value of the total complex impedance corresponds to the of the reciprocal values of the partial values: fractildeZ fracR+fractildeX_L+fractildeX_C fracR+fracjomega L+jomega CfracR-fracjomega L+jomega C fracR+jleftomega C-fracomega Lright The complex impedance is tildeZ fracfracR+jleftomega C-fracomega Lright fracfracR-jleftomega C-fracomega LrightfracR^+leftomega C-fracomega Lright^ The real impedance is given by the modulus of the complex impedance: Z sqrtRetildeZ^+ImtildeZ^fracsqrtfracR^+leftomega C-fracomega Lright^fracR^+leftomega C-fracomega Lright^ fracsqrtfracR^+leftomega C-fracomega Lright^ The phase shift corresponds to the argument: tanDeltaphi fracImtildeZRetildeZ fracfracomega L-omega CfracR Rleftfracomega L-omega Cright
Show that the impedance and phase shift of an ac circuit with a resistor resistance R a capacitor capacitance C and a coil inductance L in parallel are given by * fracZ sqrtfracR^+leftomega C-fracomega Lright^ and * tanDeltaphi Rleftfracomega L-omega Cright H: Since the voltage is the same across all three elements in the parallel circuit you can start with the corresponding phasor in a phasor diagram and add the current phasors with their respective phaseshift.
Solution:
In the phasor diagram the current phasor for the resistor is in phase with the voltage while the current phasors for the capacitor and the coil have a phase shift of mp fracpi respectively see figure. center includegraphicswidth.mm#image_path:phasors-parallel-circuit# center The phasor for the total current I can be found by adding the phasors for the partial currents I_R I_C and I_L as vectors. The amplitude of the total current is I sqrtI_R^+leftI_C-I_Lright^ Using the relations I_R fracVR I_C fracVX_C I_L fracVX_L it follows for the impedance fracZ fracIV fracsqrtleftfracVRright^+leftfracVX_C-fracVX_Lright^V sqrtfracR^+leftfracX_C-fracX_Lright^ With the expressions for the reactances X_C and X_L we find fracZomega sqrtfracR^+leftomega C-fracomega Lright^ The phase shift is given by see figure tanDeltaphi fracI_L-I_CI_R fracfracVX_L-fracVX_CfracVR RleftfracX_L-fracX_Cright Rleftfracomega L-omega Cright Alternatively the expressions can be derived using the complex reactances tildeX_L and tildeX_C. For a parallel circuit the reciprocal value of the total complex impedance corresponds to the of the reciprocal values of the partial values: fractildeZ fracR+fractildeX_L+fractildeX_C fracR+fracjomega L+jomega CfracR-fracjomega L+jomega C fracR+jleftomega C-fracomega Lright The complex impedance is tildeZ fracfracR+jleftomega C-fracomega Lright fracfracR-jleftomega C-fracomega LrightfracR^+leftomega C-fracomega Lright^ The real impedance is given by the modulus of the complex impedance: Z sqrtRetildeZ^+ImtildeZ^fracsqrtfracR^+leftomega C-fracomega Lright^fracR^+leftomega C-fracomega Lright^ fracsqrtfracR^+leftomega C-fracomega Lright^ The phase shift corresponds to the argument: tanDeltaphi fracImtildeZRetildeZ fracfracomega L-omega CfracR Rleftfracomega L-omega Cright
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Exercise:
Show that the impedance and phase shift of an ac circuit with a resistor resistance R a capacitor capacitance C and a coil inductance L in parallel are given by * fracZ sqrtfracR^+leftomega C-fracomega Lright^ and * tanDeltaphi Rleftfracomega L-omega Cright H: Since the voltage is the same across all three elements in the parallel circuit you can start with the corresponding phasor in a phasor diagram and add the current phasors with their respective phaseshift.
Solution:
In the phasor diagram the current phasor for the resistor is in phase with the voltage while the current phasors for the capacitor and the coil have a phase shift of mp fracpi respectively see figure. center includegraphicswidth.mm#image_path:phasors-parallel-circuit# center The phasor for the total current I can be found by adding the phasors for the partial currents I_R I_C and I_L as vectors. The amplitude of the total current is I sqrtI_R^+leftI_C-I_Lright^ Using the relations I_R fracVR I_C fracVX_C I_L fracVX_L it follows for the impedance fracZ fracIV fracsqrtleftfracVRright^+leftfracVX_C-fracVX_Lright^V sqrtfracR^+leftfracX_C-fracX_Lright^ With the expressions for the reactances X_C and X_L we find fracZomega sqrtfracR^+leftomega C-fracomega Lright^ The phase shift is given by see figure tanDeltaphi fracI_L-I_CI_R fracfracVX_L-fracVX_CfracVR RleftfracX_L-fracX_Cright Rleftfracomega L-omega Cright Alternatively the expressions can be derived using the complex reactances tildeX_L and tildeX_C. For a parallel circuit the reciprocal value of the total complex impedance corresponds to the of the reciprocal values of the partial values: fractildeZ fracR+fractildeX_L+fractildeX_C fracR+fracjomega L+jomega CfracR-fracjomega L+jomega C fracR+jleftomega C-fracomega Lright The complex impedance is tildeZ fracfracR+jleftomega C-fracomega Lright fracfracR-jleftomega C-fracomega LrightfracR^+leftomega C-fracomega Lright^ The real impedance is given by the modulus of the complex impedance: Z sqrtRetildeZ^+ImtildeZ^fracsqrtfracR^+leftomega C-fracomega Lright^fracR^+leftomega C-fracomega Lright^ fracsqrtfracR^+leftomega C-fracomega Lright^ The phase shift corresponds to the argument: tanDeltaphi fracImtildeZRetildeZ fracfracomega L-omega CfracR Rleftfracomega L-omega Cright
Show that the impedance and phase shift of an ac circuit with a resistor resistance R a capacitor capacitance C and a coil inductance L in parallel are given by * fracZ sqrtfracR^+leftomega C-fracomega Lright^ and * tanDeltaphi Rleftfracomega L-omega Cright H: Since the voltage is the same across all three elements in the parallel circuit you can start with the corresponding phasor in a phasor diagram and add the current phasors with their respective phaseshift.
Solution:
In the phasor diagram the current phasor for the resistor is in phase with the voltage while the current phasors for the capacitor and the coil have a phase shift of mp fracpi respectively see figure. center includegraphicswidth.mm#image_path:phasors-parallel-circuit# center The phasor for the total current I can be found by adding the phasors for the partial currents I_R I_C and I_L as vectors. The amplitude of the total current is I sqrtI_R^+leftI_C-I_Lright^ Using the relations I_R fracVR I_C fracVX_C I_L fracVX_L it follows for the impedance fracZ fracIV fracsqrtleftfracVRright^+leftfracVX_C-fracVX_Lright^V sqrtfracR^+leftfracX_C-fracX_Lright^ With the expressions for the reactances X_C and X_L we find fracZomega sqrtfracR^+leftomega C-fracomega Lright^ The phase shift is given by see figure tanDeltaphi fracI_L-I_CI_R fracfracVX_L-fracVX_CfracVR RleftfracX_L-fracX_Cright Rleftfracomega L-omega Cright Alternatively the expressions can be derived using the complex reactances tildeX_L and tildeX_C. For a parallel circuit the reciprocal value of the total complex impedance corresponds to the of the reciprocal values of the partial values: fractildeZ fracR+fractildeX_L+fractildeX_C fracR+fracjomega L+jomega CfracR-fracjomega L+jomega C fracR+jleftomega C-fracomega Lright The complex impedance is tildeZ fracfracR+jleftomega C-fracomega Lright fracfracR-jleftomega C-fracomega LrightfracR^+leftomega C-fracomega Lright^ The real impedance is given by the modulus of the complex impedance: Z sqrtRetildeZ^+ImtildeZ^fracsqrtfracR^+leftomega C-fracomega Lright^fracR^+leftomega C-fracomega Lright^ fracsqrtfracR^+leftomega C-fracomega Lright^ The phase shift corresponds to the argument: tanDeltaphi fracImtildeZRetildeZ fracfracomega L-omega CfracR Rleftfracomega L-omega Cright
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