Resonance in LCR Series Circuit
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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?
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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
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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
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Exercise:
In an LCR series circuit the three partial voltages have their maxima at slightly different frequencies. abcliste abc Derive the expressions for the partial voltages and the angular frequencies at which they reach their respective maximum. abc Graph the partial voltages vs. the angular frequency and calculate the numerical values for the resonance frequencies for a circuit with RO resistance CO capacitance and LO inductance. abc What are the mechanical equivalents for the partial voltages? abcliste
Solution:
blaufabcliste abc The voltage across the capacitor can be expressed as V_C X_C I X_C fracV_Z fracomega CfracV_sqrtR^+leftomega L-/omega Cright^ fracV_LCsqrtR/L^ omega^+leftomega^-/LCright^ Using the expression for the angular frequency omega_ and the damping constant delta this can be written as V_C fracV_/omega_^ sqrtdelta^omega^+leftomega^-omega_^right^ V_ fracomega_^sqrtdelta^omega^+leftomega^-omega_^right^ labelvc The capacitor voltage has its maximum amplitude if the denominator in refvc has its minimum. This is the case if dvleftdelta^omega^+leftomega^-omega_^right^rightomega delta^omega+leftomega^-omega_^right omega omegaleftdelta^+omega^-omega_^right It follows that either omega in which case it can be shown that the capacitor voltage has a local minimum or omega^omega_^-delta^. The second solution corresponds to the em resonance frequency for the capacitor voltage: omega_RC omCF vspacemm The resistor voltage is given by V_Romega R I RfracV_Z RfracV_sqrtR^+leftomega L-/omega Cright^ We have seen earlier that this function has a maximum for omega_RR omega_ omRF so this is the resonance frequency for the resistor voltage. vspacemm The inductor voltage is given by V_L X_L I X_L fracV_Z fracomega L V_sqrtR^+leftomega L-/omega Cright^ fracomega V_sqrtR/L^+leftomega-/omega L Cright^ fracomega^ V_sqrtdelta^omega^+leftomega^-omega_^right^ The condition for a maximum is dvV_Lomegaomega fracomega V_sqrtdelta^omega^+leftomega^-omega_^right^ &quadquad -fracfracomega^ V_leftdelta^omega^+leftomega^-omega_^right^right^/ leftdelta^omega+leftomega^-omega_^right omegaright V_fracomegaleftdelta^omega^+leftomega^-omega_^right^right-omega^leftdelta^omega+omegaleftomega^-omega_^rightrightleftdelta^omega^+leftomega^-omega_^right^right^/ For omeganeq the condition can be simplified as delta^omega^+omega^-omega^omega_^+omega_^-delta^omega^-omega^+omega^omega_^ delta^omega^-omega^omega_^+omega_^ omega^leftdelta^-omega_^right+omega_^ The positiv solution for omega is the resonance frequency for the inductor voltage: omega_RL omLF vspacemm em Remark: The three different resonance frequencies are related via the relation omega_RR sqrtomega_RComega_RL i.e. the resonance frequency for the resistor voltage is the geometric mean of the other two resonance frequencies. abc The angular frequency of the undamped oscillation which is also the resonance frequency for the resistor voltage is omega_RR omega_ omRF fracsqrtLtimesC resultomRS The damping constant is delta deF fracRtimesL deS The resonance frequencies for the capacitor and inductor voltage are omega_RC omCF sqrtomR^-timesde^ resultomCS omega_RL omLF fracomR^sqrtomR^-timesde^ resultomLS The graphs of the three partial voltages are shown in the figure below. For the angular frequency omega_ peak of the resistor voltage the capacitor and the inductor voltage have the same amplitudes. Since they are in opposite phase it is obvious that they cancel each other so the total voltage is equal to the resistor voltage. This is a special case of the general result that the three partial voltages add up to the input voltage but because of the phase shift between the partial voltages this is not obvious for most frequencies. center includegraphicswidthcm#image_path:lcr-series-resonance# center abc The displacement yt of a mechanical oscillation corresponds to the charge qt in the LCR oscillator. It follows that the velocity corresponds to first derivative dot qt and that the acceleration corresponds to the second derivative ddot qt. Using the equivalence between mechanical and electrical oscillators i.e. m & leftrightarrow L k & leftrightarrow fracC beta & leftrightarrow R we find the following correspondences: v_Ct fracqtC leftrightarrow F_kt k yt v_Rt R it R dot qt leftrightarrow F_dt beta v_yt v_Lt -Lddot imatht leftrightarrow F_mt -m a_yt Each voltage corresponds to a force in the mechanical system: The capacitor voltage corresponds to the elastic force of the spring the resistor voltage to the drag force and the inductor voltage to the em erial force. Additionaly the input voltage corresponds to a driving force Ft F_cosomega t Putting everything together we find that F_mt m a_yt Ft-F_kt-F_dt which is Newton's second law for the mechanical system. abcliste
In an LCR series circuit the three partial voltages have their maxima at slightly different frequencies. abcliste abc Derive the expressions for the partial voltages and the angular frequencies at which they reach their respective maximum. abc Graph the partial voltages vs. the angular frequency and calculate the numerical values for the resonance frequencies for a circuit with RO resistance CO capacitance and LO inductance. abc What are the mechanical equivalents for the partial voltages? abcliste
Solution:
blaufabcliste abc The voltage across the capacitor can be expressed as V_C X_C I X_C fracV_Z fracomega CfracV_sqrtR^+leftomega L-/omega Cright^ fracV_LCsqrtR/L^ omega^+leftomega^-/LCright^ Using the expression for the angular frequency omega_ and the damping constant delta this can be written as V_C fracV_/omega_^ sqrtdelta^omega^+leftomega^-omega_^right^ V_ fracomega_^sqrtdelta^omega^+leftomega^-omega_^right^ labelvc The capacitor voltage has its maximum amplitude if the denominator in refvc has its minimum. This is the case if dvleftdelta^omega^+leftomega^-omega_^right^rightomega delta^omega+leftomega^-omega_^right omega omegaleftdelta^+omega^-omega_^right It follows that either omega in which case it can be shown that the capacitor voltage has a local minimum or omega^omega_^-delta^. The second solution corresponds to the em resonance frequency for the capacitor voltage: omega_RC omCF vspacemm The resistor voltage is given by V_Romega R I RfracV_Z RfracV_sqrtR^+leftomega L-/omega Cright^ We have seen earlier that this function has a maximum for omega_RR omega_ omRF so this is the resonance frequency for the resistor voltage. vspacemm The inductor voltage is given by V_L X_L I X_L fracV_Z fracomega L V_sqrtR^+leftomega L-/omega Cright^ fracomega V_sqrtR/L^+leftomega-/omega L Cright^ fracomega^ V_sqrtdelta^omega^+leftomega^-omega_^right^ The condition for a maximum is dvV_Lomegaomega fracomega V_sqrtdelta^omega^+leftomega^-omega_^right^ &quadquad -fracfracomega^ V_leftdelta^omega^+leftomega^-omega_^right^right^/ leftdelta^omega+leftomega^-omega_^right omegaright V_fracomegaleftdelta^omega^+leftomega^-omega_^right^right-omega^leftdelta^omega+omegaleftomega^-omega_^rightrightleftdelta^omega^+leftomega^-omega_^right^right^/ For omeganeq the condition can be simplified as delta^omega^+omega^-omega^omega_^+omega_^-delta^omega^-omega^+omega^omega_^ delta^omega^-omega^omega_^+omega_^ omega^leftdelta^-omega_^right+omega_^ The positiv solution for omega is the resonance frequency for the inductor voltage: omega_RL omLF vspacemm em Remark: The three different resonance frequencies are related via the relation omega_RR sqrtomega_RComega_RL i.e. the resonance frequency for the resistor voltage is the geometric mean of the other two resonance frequencies. abc The angular frequency of the undamped oscillation which is also the resonance frequency for the resistor voltage is omega_RR omega_ omRF fracsqrtLtimesC resultomRS The damping constant is delta deF fracRtimesL deS The resonance frequencies for the capacitor and inductor voltage are omega_RC omCF sqrtomR^-timesde^ resultomCS omega_RL omLF fracomR^sqrtomR^-timesde^ resultomLS The graphs of the three partial voltages are shown in the figure below. For the angular frequency omega_ peak of the resistor voltage the capacitor and the inductor voltage have the same amplitudes. Since they are in opposite phase it is obvious that they cancel each other so the total voltage is equal to the resistor voltage. This is a special case of the general result that the three partial voltages add up to the input voltage but because of the phase shift between the partial voltages this is not obvious for most frequencies. center includegraphicswidthcm#image_path:lcr-series-resonance# center abc The displacement yt of a mechanical oscillation corresponds to the charge qt in the LCR oscillator. It follows that the velocity corresponds to first derivative dot qt and that the acceleration corresponds to the second derivative ddot qt. Using the equivalence between mechanical and electrical oscillators i.e. m & leftrightarrow L k & leftrightarrow fracC beta & leftrightarrow R we find the following correspondences: v_Ct fracqtC leftrightarrow F_kt k yt v_Rt R it R dot qt leftrightarrow F_dt beta v_yt v_Lt -Lddot imatht leftrightarrow F_mt -m a_yt Each voltage corresponds to a force in the mechanical system: The capacitor voltage corresponds to the elastic force of the spring the resistor voltage to the drag force and the inductor voltage to the em erial force. Additionaly the input voltage corresponds to a driving force Ft F_cosomega t Putting everything together we find that F_mt m a_yt Ft-F_kt-F_dt which is Newton's second law for the mechanical system. abcliste
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Exercise:
In an LCR series circuit the three partial voltages have their maxima at slightly different frequencies. abcliste abc Derive the expressions for the partial voltages and the angular frequencies at which they reach their respective maximum. abc Graph the partial voltages vs. the angular frequency and calculate the numerical values for the resonance frequencies for a circuit with RO resistance CO capacitance and LO inductance. abc What are the mechanical equivalents for the partial voltages? abcliste
Solution:
blaufabcliste abc The voltage across the capacitor can be expressed as V_C X_C I X_C fracV_Z fracomega CfracV_sqrtR^+leftomega L-/omega Cright^ fracV_LCsqrtR/L^ omega^+leftomega^-/LCright^ Using the expression for the angular frequency omega_ and the damping constant delta this can be written as V_C fracV_/omega_^ sqrtdelta^omega^+leftomega^-omega_^right^ V_ fracomega_^sqrtdelta^omega^+leftomega^-omega_^right^ labelvc The capacitor voltage has its maximum amplitude if the denominator in refvc has its minimum. This is the case if dvleftdelta^omega^+leftomega^-omega_^right^rightomega delta^omega+leftomega^-omega_^right omega omegaleftdelta^+omega^-omega_^right It follows that either omega in which case it can be shown that the capacitor voltage has a local minimum or omega^omega_^-delta^. The second solution corresponds to the em resonance frequency for the capacitor voltage: omega_RC omCF vspacemm The resistor voltage is given by V_Romega R I RfracV_Z RfracV_sqrtR^+leftomega L-/omega Cright^ We have seen earlier that this function has a maximum for omega_RR omega_ omRF so this is the resonance frequency for the resistor voltage. vspacemm The inductor voltage is given by V_L X_L I X_L fracV_Z fracomega L V_sqrtR^+leftomega L-/omega Cright^ fracomega V_sqrtR/L^+leftomega-/omega L Cright^ fracomega^ V_sqrtdelta^omega^+leftomega^-omega_^right^ The condition for a maximum is dvV_Lomegaomega fracomega V_sqrtdelta^omega^+leftomega^-omega_^right^ &quadquad -fracfracomega^ V_leftdelta^omega^+leftomega^-omega_^right^right^/ leftdelta^omega+leftomega^-omega_^right omegaright V_fracomegaleftdelta^omega^+leftomega^-omega_^right^right-omega^leftdelta^omega+omegaleftomega^-omega_^rightrightleftdelta^omega^+leftomega^-omega_^right^right^/ For omeganeq the condition can be simplified as delta^omega^+omega^-omega^omega_^+omega_^-delta^omega^-omega^+omega^omega_^ delta^omega^-omega^omega_^+omega_^ omega^leftdelta^-omega_^right+omega_^ The positiv solution for omega is the resonance frequency for the inductor voltage: omega_RL omLF vspacemm em Remark: The three different resonance frequencies are related via the relation omega_RR sqrtomega_RComega_RL i.e. the resonance frequency for the resistor voltage is the geometric mean of the other two resonance frequencies. abc The angular frequency of the undamped oscillation which is also the resonance frequency for the resistor voltage is omega_RR omega_ omRF fracsqrtLtimesC resultomRS The damping constant is delta deF fracRtimesL deS The resonance frequencies for the capacitor and inductor voltage are omega_RC omCF sqrtomR^-timesde^ resultomCS omega_RL omLF fracomR^sqrtomR^-timesde^ resultomLS The graphs of the three partial voltages are shown in the figure below. For the angular frequency omega_ peak of the resistor voltage the capacitor and the inductor voltage have the same amplitudes. Since they are in opposite phase it is obvious that they cancel each other so the total voltage is equal to the resistor voltage. This is a special case of the general result that the three partial voltages add up to the input voltage but because of the phase shift between the partial voltages this is not obvious for most frequencies. center includegraphicswidthcm#image_path:lcr-series-resonance# center abc The displacement yt of a mechanical oscillation corresponds to the charge qt in the LCR oscillator. It follows that the velocity corresponds to first derivative dot qt and that the acceleration corresponds to the second derivative ddot qt. Using the equivalence between mechanical and electrical oscillators i.e. m & leftrightarrow L k & leftrightarrow fracC beta & leftrightarrow R we find the following correspondences: v_Ct fracqtC leftrightarrow F_kt k yt v_Rt R it R dot qt leftrightarrow F_dt beta v_yt v_Lt -Lddot imatht leftrightarrow F_mt -m a_yt Each voltage corresponds to a force in the mechanical system: The capacitor voltage corresponds to the elastic force of the spring the resistor voltage to the drag force and the inductor voltage to the em erial force. Additionaly the input voltage corresponds to a driving force Ft F_cosomega t Putting everything together we find that F_mt m a_yt Ft-F_kt-F_dt which is Newton's second law for the mechanical system. abcliste
In an LCR series circuit the three partial voltages have their maxima at slightly different frequencies. abcliste abc Derive the expressions for the partial voltages and the angular frequencies at which they reach their respective maximum. abc Graph the partial voltages vs. the angular frequency and calculate the numerical values for the resonance frequencies for a circuit with RO resistance CO capacitance and LO inductance. abc What are the mechanical equivalents for the partial voltages? abcliste
Solution:
blaufabcliste abc The voltage across the capacitor can be expressed as V_C X_C I X_C fracV_Z fracomega CfracV_sqrtR^+leftomega L-/omega Cright^ fracV_LCsqrtR/L^ omega^+leftomega^-/LCright^ Using the expression for the angular frequency omega_ and the damping constant delta this can be written as V_C fracV_/omega_^ sqrtdelta^omega^+leftomega^-omega_^right^ V_ fracomega_^sqrtdelta^omega^+leftomega^-omega_^right^ labelvc The capacitor voltage has its maximum amplitude if the denominator in refvc has its minimum. This is the case if dvleftdelta^omega^+leftomega^-omega_^right^rightomega delta^omega+leftomega^-omega_^right omega omegaleftdelta^+omega^-omega_^right It follows that either omega in which case it can be shown that the capacitor voltage has a local minimum or omega^omega_^-delta^. The second solution corresponds to the em resonance frequency for the capacitor voltage: omega_RC omCF vspacemm The resistor voltage is given by V_Romega R I RfracV_Z RfracV_sqrtR^+leftomega L-/omega Cright^ We have seen earlier that this function has a maximum for omega_RR omega_ omRF so this is the resonance frequency for the resistor voltage. vspacemm The inductor voltage is given by V_L X_L I X_L fracV_Z fracomega L V_sqrtR^+leftomega L-/omega Cright^ fracomega V_sqrtR/L^+leftomega-/omega L Cright^ fracomega^ V_sqrtdelta^omega^+leftomega^-omega_^right^ The condition for a maximum is dvV_Lomegaomega fracomega V_sqrtdelta^omega^+leftomega^-omega_^right^ &quadquad -fracfracomega^ V_leftdelta^omega^+leftomega^-omega_^right^right^/ leftdelta^omega+leftomega^-omega_^right omegaright V_fracomegaleftdelta^omega^+leftomega^-omega_^right^right-omega^leftdelta^omega+omegaleftomega^-omega_^rightrightleftdelta^omega^+leftomega^-omega_^right^right^/ For omeganeq the condition can be simplified as delta^omega^+omega^-omega^omega_^+omega_^-delta^omega^-omega^+omega^omega_^ delta^omega^-omega^omega_^+omega_^ omega^leftdelta^-omega_^right+omega_^ The positiv solution for omega is the resonance frequency for the inductor voltage: omega_RL omLF vspacemm em Remark: The three different resonance frequencies are related via the relation omega_RR sqrtomega_RComega_RL i.e. the resonance frequency for the resistor voltage is the geometric mean of the other two resonance frequencies. abc The angular frequency of the undamped oscillation which is also the resonance frequency for the resistor voltage is omega_RR omega_ omRF fracsqrtLtimesC resultomRS The damping constant is delta deF fracRtimesL deS The resonance frequencies for the capacitor and inductor voltage are omega_RC omCF sqrtomR^-timesde^ resultomCS omega_RL omLF fracomR^sqrtomR^-timesde^ resultomLS The graphs of the three partial voltages are shown in the figure below. For the angular frequency omega_ peak of the resistor voltage the capacitor and the inductor voltage have the same amplitudes. Since they are in opposite phase it is obvious that they cancel each other so the total voltage is equal to the resistor voltage. This is a special case of the general result that the three partial voltages add up to the input voltage but because of the phase shift between the partial voltages this is not obvious for most frequencies. center includegraphicswidthcm#image_path:lcr-series-resonance# center abc The displacement yt of a mechanical oscillation corresponds to the charge qt in the LCR oscillator. It follows that the velocity corresponds to first derivative dot qt and that the acceleration corresponds to the second derivative ddot qt. Using the equivalence between mechanical and electrical oscillators i.e. m & leftrightarrow L k & leftrightarrow fracC beta & leftrightarrow R we find the following correspondences: v_Ct fracqtC leftrightarrow F_kt k yt v_Rt R it R dot qt leftrightarrow F_dt beta v_yt v_Lt -Lddot imatht leftrightarrow F_mt -m a_yt Each voltage corresponds to a force in the mechanical system: The capacitor voltage corresponds to the elastic force of the spring the resistor voltage to the drag force and the inductor voltage to the em erial force. Additionaly the input voltage corresponds to a driving force Ft F_cosomega t Putting everything together we find that F_mt m a_yt Ft-F_kt-F_dt which is Newton's second law for the mechanical system. abcliste
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