SOLUTION: A capacitor was connected in series with a resistor and then charged up from a power supply. The voltage across the capacitor was measured on an oscilloscope and then compared with

Algebra ->  Rational-functions -> SOLUTION: A capacitor was connected in series with a resistor and then charged up from a power supply. The voltage across the capacitor was measured on an oscilloscope and then compared with      Log On


   

Question 1182991: A capacitor was connected in series with a resistor and then charged up from a power supply. The voltage across the capacitor was measured on an oscilloscope and then compared with those when calculated.
Circuit information:
Capacitor = 80 nF (80x10-9 F) Resistor = 40 kΩ (40x103 Ω) Supply voltage, V = 5 V
Charging voltage v = V(1-e^((-t)/T) ), where T = CR and is called the time constant.
Calculate the time constant for the circuit
Use a spreadsheet to plot a curve of the charging circuit over the range of 0 to 20 ms(milliseconds).
Use differentiation to find the rate of change of voltage at 6 ms
From your plotted graph measure the gradient using the normal method to determine the gradient of a curve at a point where the time is 6 ms.
Compare the answers found in (iii) and (iv)
A discharging circuit has voltage across the capacitor of v = V e^((-t)/RC)
Determine dv/dt , the rate of change of voltage in the discharging circuit capacitor. Then find dv/dt when t = T
Answer by Solver92311(821) About Me  (Show Source):
You can put this solution on YOUR website!











where is elapsed time in seconds. 

(msec)    1       2       3       4       5       6       7       8       9       10
        1.342	2.324	3.042	3.567	3.952	4.233	4.439	4.590	4.700   4.780	

          11      12      13      14      15      16      17      18      19      20 
	4.839	4.882	4.914	4.937	4.954	4.966	4.975	4.982	4.987	4.990



I don't have a way to measure the gradient at a point on an excel graph, so I approximated by calculating and this differs from the value of the first derivative at 0.006 by only one digit in the fourth decimal place.











Essentially, zero.

John

My calculator said it, I believe it, that settles it

From
I > Ø