i ALREADY RE-SUBMITTED THE QUESTIONS AND IT WAS CHARGED AS QUESTION #2 WHICH SUP
ID: 2080096 • Letter: I
Question
i ALREADY RE-SUBMITTED THE QUESTIONS AND IT WAS CHARGED AS QUESTION #2 WHICH SUPPOSED TO BE ONLY QUESTION #01 BECAUSE YOU ASKED FOR A CLARIFICATION/ RESUBMISSION OF THE QUESTION....
You used the following circuit in your experiments? What is this circuit designed to measure? What does a reading of 10 Volts on the meter indicate? Be specific. What is the function of PC_2? What is the purpose or R_2? What is the function of L_1? Why does the voltage peak and they immediately go down? If the value of C_2 were 3000 mu F, what would a reading of 10 V immediately after pressing the button indicated?Explanation / Answer
NB: From the circuit, the input seems to be a DC source (The figure is not so clear). However, I am also providing the answers to the above questions if the input would have been ac. ie, The answers are given correspondingly for dc and ac.
30. The circuit is designed to measure the capacitor voltage. (If the input voltage would have been ac,then the circuit will measure the peak voltage because the capacitor will be charged upto the peak voltage, ie, the circuit would be a peak detector circuit).
31. Irrespective of then nature of the input voltage, 10V reading of the voltmeter indicates that the instantaneous voltage across the capacitor is 10V.
32. The PB2, when active, makes sure that the voltage across the capacitor is less that 20V so that the voltmeter can safely operate. Also when the source is de-activated, the PB2 provides a discharging path for the capacitor through R2.
33. R2 will act as a load to the capacitor while discharging. (If the input would have been ac, the capacitor discharges through R2 during the negative half cycle.)
34. D1 makes sure that no negative voltage or negative noise spike get impressed across the capacitor and thereby the capacitor voltage will be always positive. Hence D1 also allows the voltmeter to operate safely.
The role of D1 is more pronounced or evident if the input source would have been ac as it contains posuitive as well as negative half cycle.
35. When the PB2 is active, R2 acts as a load and hence the capacitor discharges through R2. One more way to interpret the discharge activity is through the circuit time constant.
When PB2 is active, the time constant, T is given by,
T= R2C = 30*1000*10-6 = 30mSec
As the time constant is very small, the voltage immediately goes down. To be specific, the capacitor voltage will degrade to 0.2V at 150mSec after the button is pressed.
If the input source would have been ac, the capacitor voltage will goes to its peak during the positive half cycle and then will discharge slowly through R2 during negative cycle of the input voltage.
36. If the capacitor value is increased to 3000 micro Farad, the time constant, T will increase to 90mSec. Thus the capacitor voltage degrades at a lower rate. However, as per the property of the capacitor, ie, capacitor doesnot allow sudden change in its voltage, we can say that the capacitor voltage will be unchanged and will be exactly 10V at the very next instant after the button is pressed. The capacitor voltage will be degraded to almost 0.2V at 450mSec after the button is pressed.
(If the input source would have been ac, the capacitor voltage won't go down immediately during the negative half cycle if the capacitor value is increased.)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.