Capacitor discharge/charge
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH04_01_que_20120307.pdf
For question 16c of the paper above why is it important for the capacitor to fully charge or discharge before sound comes in? Does the sound break the circuit or something? Mark scheme below.
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH04_01_msc_20120208.pdf
For question 16c of the paper above why is it important for the capacitor to fully charge or discharge before sound comes in? Does the sound break the circuit or something? Mark scheme below.
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH04_01_msc_20120208.pdf
[QUOTE="runny4;65521991"]http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH04_01_que_20120307.pdf
For question 16c of the paper above why is it important for the capacitor to fully charge or discharge before sound comes in? Does the sound break the circuit or something? Mark scheme below.
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH04_01_msc_20120208.pdf[/QUOTE]
The mark scheme answer does not state 'before the sound comes in'.
Fully charging and discharging is required to produce the greatest p.d. across the resistor from which the a.c. transcription of the incoming sound is derived.
Maximum charging and discharging allows the greatest output voltage signal (loudest sound w.r.t. no sound) to be produced. This in turn improves the signal to noise ratio of the signal since the microphone output is of a very low level (mV typically) and susceptible to external electromagnetic interference from unwanted sources such as 50Hz electricity mains interference which would be superimposed on top of the wanted signal.
For question 16c of the paper above why is it important for the capacitor to fully charge or discharge before sound comes in? Does the sound break the circuit or something? Mark scheme below.
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH04_01_msc_20120208.pdf[/QUOTE]
The mark scheme answer does not state 'before the sound comes in'.
Fully charging and discharging is required to produce the greatest p.d. across the resistor from which the a.c. transcription of the incoming sound is derived.
Maximum charging and discharging allows the greatest output voltage signal (loudest sound w.r.t. no sound) to be produced. This in turn improves the signal to noise ratio of the signal since the microphone output is of a very low level (mV typically) and susceptible to external electromagnetic interference from unwanted sources such as 50Hz electricity mains interference which would be superimposed on top of the wanted signal.
[QUOTE="uberteknik;65534975"]
Then for part b), why is it an alternating pd- how can it be +ve and -ve, and for part c i don't understand how the switch in the circuit works- does it randomly switch on and off?
Original post by runny4
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH04_01_que_20120307.pdf
For question 16c of the paper above why is it important for the capacitor to fully charge or discharge before sound comes in? Does the sound break the circuit or something? Mark scheme below.
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH04_01_msc_20120208.pdf[/QUOTE]
The mark scheme answer does not state 'before the sound comes in'.
Fully charging and discharging is required to produce the greatest p.d. across the resistor from which the a.c. transcription of the incoming sound is derived.
Maximum charging and discharging allows the greatest output voltage signal (loudest sound w.r.t. no sound) to be produced. This in turn improves the signal to noise ratio of the signal since the microphone output is of a very low level (mV typically) and susceptible to external electromagnetic interference from unwanted sources such as 50Hz electricity mains interference which would be superimposed on top of the wanted signal.
For question 16c of the paper above why is it important for the capacitor to fully charge or discharge before sound comes in? Does the sound break the circuit or something? Mark scheme below.
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH04_01_msc_20120208.pdf[/QUOTE]
The mark scheme answer does not state 'before the sound comes in'.
Fully charging and discharging is required to produce the greatest p.d. across the resistor from which the a.c. transcription of the incoming sound is derived.
Maximum charging and discharging allows the greatest output voltage signal (loudest sound w.r.t. no sound) to be produced. This in turn improves the signal to noise ratio of the signal since the microphone output is of a very low level (mV typically) and susceptible to external electromagnetic interference from unwanted sources such as 50Hz electricity mains interference which would be superimposed on top of the wanted signal.
Then for part b), why is it an alternating pd- how can it be +ve and -ve, and for part c i don't understand how the switch in the circuit works- does it randomly switch on and off?
[QUOTE="uberteknik;65534975"]
please reply
Original post by runny4
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH04_01_que_20120307.pdf
For question 16c of the paper above why is it important for the capacitor to fully charge or discharge before sound comes in? Does the sound break the circuit or something? Mark scheme below.
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH04_01_msc_20120208.pdf[/QUOTE]
The mark scheme answer does not state 'before the sound comes in'.
Fully charging and discharging is required to produce the greatest p.d. across the resistor from which the a.c. transcription of the incoming sound is derived.
Maximum charging and discharging allows the greatest output voltage signal (loudest sound w.r.t. no sound) to be produced. This in turn improves the signal to noise ratio of the signal since the microphone output is of a very low level (mV typically) and susceptible to external electromagnetic interference from unwanted sources such as 50Hz electricity mains interference which would be superimposed on top of the wanted signal.
For question 16c of the paper above why is it important for the capacitor to fully charge or discharge before sound comes in? Does the sound break the circuit or something? Mark scheme below.
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH04_01_msc_20120208.pdf[/QUOTE]
The mark scheme answer does not state 'before the sound comes in'.
Fully charging and discharging is required to produce the greatest p.d. across the resistor from which the a.c. transcription of the incoming sound is derived.
Maximum charging and discharging allows the greatest output voltage signal (loudest sound w.r.t. no sound) to be produced. This in turn improves the signal to noise ratio of the signal since the microphone output is of a very low level (mV typically) and susceptible to external electromagnetic interference from unwanted sources such as 50Hz electricity mains interference which would be superimposed on top of the wanted signal.
please reply
[QUOTE="runny4;65537377"]
Q = CV
V = Q/C
Capacitance is a function of the plate area and the distance (dielectric gap) between the plates.
Changes in air pressure, causes the distance between the capacitor plates to alter. i.e. the plate distance 'd', changes as a function of the sound pressure waves hitting the non-fixed plate.
Thus the capacitance alters dynamically as a facsimile of the sound.
Therefore the charge stored on the plates must also change as a facsimile of the sound pressure.
The a.c. waveform is a result of the movement of charge around the closed circuit as the electric force between the capacitor plates constantly equalises.
Since charge flow is via the resistor, a time varying p.d. across the resistor is developed in response to the sound pressure changes.
Original post by uberteknik
Then for part b), why is it an alternating pd- how can it be +ve and -ve, and for part c i don't understand how the switch in the circuit works- does it randomly switch on and off?
Then for part b), why is it an alternating pd- how can it be +ve and -ve, and for part c i don't understand how the switch in the circuit works- does it randomly switch on and off?
Q = CV
V = Q/C
Capacitance is a function of the plate area and the distance (dielectric gap) between the plates.
Changes in air pressure, causes the distance between the capacitor plates to alter. i.e. the plate distance 'd', changes as a function of the sound pressure waves hitting the non-fixed plate.
Thus the capacitance alters dynamically as a facsimile of the sound.
Therefore the charge stored on the plates must also change as a facsimile of the sound pressure.
The a.c. waveform is a result of the movement of charge around the closed circuit as the electric force between the capacitor plates constantly equalises.
Since charge flow is via the resistor, a time varying p.d. across the resistor is developed in response to the sound pressure changes.
(edited 7 years ago)
[QUOTE="runny4;65618273"]My response is clear. Do you now understand how a capacitor can be used as a microphone?
Original post by uberteknik
please reply
please reply
[QUOTE="uberteknik;65667611"]
yes thank you
Original post by runny4
My response is clear. Do you now understand how a capacitor can be used as a microphone?
yes thank you
Quick Reply
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