Question: Capacitors, Time Constant and p.d.

In summary: For d) and e) you need to get the discharge curve V_C vs t from the data givenand find the time constant, \tau, from the graph.Welcome to PF ProD.In summary, the conversation discusses the discharge of a capacitor connected to a digital voltmeter and a resistor. The results of the experiment were recorded and used to determine the current in the resistor at 30 seconds, plot a graph of p.d. against time, calculate the capacitance of the capacitor, demonstrate the exponential fall of p.d., and calculate a second value for the capacitance using the time constant. The reliability of using different methods to find the capacitance is also discussed.
  • #1
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if anyone could help me on this it would be much apreciated, thanks :smile: .

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
 
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  • #2
Welcome to PF ProD.

a) is OK.

For c):

According to the theory the discharge voltage decays exponentially

[tex]V_C = V_o e^{-\frac{t}{\tau}}[/tex]

which means that the gradient of the graph will be

[tex]grad \ =-\frac{1}{\tau} V_C[/tex]

so the positive inverse of the gradient gives the capacitive time constant, [tex]\tau = RC[/tex], of the circuit - we assume that [tex]V_C = 1.0\ V[/tex] at 30 seconds.
 
Last edited:
  • #3
for any help you can provide!



a) Yes, you are correct. The current can be calculated using Ohm's law as I=V/R where V is the voltage and R is the resistance.

b) Yes, to measure the rate of decrease of p.d. immediately after t=30s, you would need to calculate the gradient of the graph. The gradient represents the change in p.d. (y-axis) over time (x-axis). In this case, the change in p.d. would be from t=30s to the next data point at t=40s.

c) To calculate the capacitance, you would need to use the formula C=Q/V. As you mentioned, the charge (Q) is not known, so you can substitute Q=IT. You can use the values of current and time that you calculated in part (a). The voltage (V) would be the initial p.d. of 1V.

d)
i) Yes, you can plot a graph of ln(p.d.) against time (t) to demonstrate the exponential decay of p.d. This would result in a straight line with a negative slope.

ii) To deduce the time constant, you would need to use the formula for the time constant, which is RC. In this case, R is the resistance of 1.5 mega ohms and C is the capacitance that you calculated in part (c).

e) To calculate a second value for the capacitance, you can use the formula C=RC/t where R is the resistance and t is the time constant that you calculated in part (d). This would give you another value for the capacitance.

f) Both methods of finding the capacitance can give reliable values. However, using the time constant method may be more accurate as it takes into account the resistance of the circuit. The method in part (c) assumes that the resistance is zero, which may not be the case in a real circuit.
 

Related to Question: Capacitors, Time Constant and p.d.

1. What is a capacitor?

A capacitor is an electronic component that stores electrical energy. It is made up of two conductive plates separated by an insulating material, and when connected to a power source, it can hold and release a charge.

2. What is the time constant of a capacitor?

The time constant of a capacitor is the amount of time it takes for the capacitor to charge up to about 63% of its maximum capacity or discharge to about 37% of its initial charge. It is calculated by multiplying the capacitance (C) of the capacitor with the resistance (R) in the circuit, or τ = RC.

3. How does a capacitor affect the voltage in a circuit?

A capacitor can affect the voltage in a circuit by storing and releasing electrical energy. When a capacitor is connected to a power source, it charges up and creates a potential difference (p.d.) between its plates. This p.d. decreases as the capacitor discharges, and it can also affect the overall voltage in the circuit.

4. What is the relationship between capacitance and p.d. in a capacitor?

The relationship between capacitance and p.d. in a capacitor is inversely proportional. This means that as the capacitance increases, the p.d. decreases, and vice versa. This relationship is described by the formula Q = CV, where Q is the charge stored, C is the capacitance, and V is the p.d.

5. How are capacitors used in electronic circuits?

Capacitors are used in electronic circuits for a variety of purposes. They can be used to store energy, filter out unwanted frequencies, stabilize voltage, and even act as timing elements. They are commonly found in power supply circuits, audio equipment, and electronic devices like computers and smartphones.

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