Question: Capacitors, Time Constant and p.d.

[ProD]

if anyone could help me on this it would be much apreciated, thanks .

Question:
A capacitor was joined across a digital voltmeter and charged to a p.d. of 1V. the p.d. was then measured at 20s intervals and at t=30s a resistor (R=1.5 mega ohms (1500000ohms)) was connected across the capacitor.
The following results were obtained:

time (s) 0.00 * 20.0 * 40.0 * 60.0 * 80.0 * 100.0 * 120.0
p.d. (V) 1.00 * 1.00 * 0.81 * 0.54 * 0.35 * 0.23 * 0.15

a) what is the current in the resistor at t=30s?

b) plot a graph of p.d. against time and measure the rate of decrease of p.d immediately after t=30s

c) hence calculate the capacitance of the capacitor

d) i) plot a graph of ln(p.d.) against timine (t) to demonstrate the ecponential fall of p.d.

ii) deduce the time constant of the decay process.

e) use your value of the time constant to calculate a second value for the capacitance of the capacitor.

f) explain which method of finding C gives the more reliable value.


My thought:

a) V=IR
I=V/R
I=1/1500000
I=0.000000667 Amps or (6.67Exp-7)

b) *drawn graph on paper* for the rate of decreasing p.d. do i take the gradient of the graph after 30s?
dy/dx... in this case...dV/dt

c) C=Q/V

Charge (Q) is not know so substitute Q=IT

therefore: C=IT/V

what values do i take for the time and voltage, do i use the ones i used to find the current in part (a)..?

d)
i) *done on paper*

ii) time constant = RC
R=1.5Exp6 ohms
C= value found in part (c)??

e) ?

f) ?






Thanks in anticipation

 

[andrevdh]

Welcome to PF ProD.

a) is OK.

For c):

According to the theory the discharge voltage decays exponentially

$$V_C = V_o e^{-\frac{t}{\tau}}$$

which means that the gradient of the graph will be

$$grad \ =-\frac{1}{\tau} V_C$$

so the positive inverse of the gradient gives the capacitive time constant, $$\tau = RC$$, of the circuit - we assume that $$V_C = 1.0\ V$$ at 30 seconds.