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

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) (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