Episode 129: Discharge of a capacitor: Q = Qoe-t/CR
Students will have already seen that the
discharge is not a steady process (Episode 125), but it is useful to have
graphical evidence before discussing the theory.
You need to build up your students’ understanding
of exponential processes, through experiments, and through graphical and
algebraic approaches, all related to the underlying physical processes
involved. For the more mathematically able students, you may even be able to
use calculus.
This episode is a long one, and may spread
over several teaching sessions.
An Excel spreadsheet is included as part of
the Student Questions for activity 129-7
Summary
Student
experiment: Exponential discharge. (30 minutes)
Discussion:
Characteristics of exponentials. (20 minutes)
Student
activity: Spreadsheet model. (20 minutes)
Student
experiment: Varying R and C. (30 minutes)
Discussion:
Deriving exponential equations. (30 minutes)
Worked
example: Using the equations. (20 minutes)
Discussion:
Time constant. (15 minutes)
Student
activity: Analysing graphs. (15 minutes)
Student
questions: Practice with equations. (30 minutes)
Discussion:
Back to reality. (10 minutes)
Student experiment:
Exponential discharge
The suggestion to look for a pattern by
measuring halving times is worth pursuing. It forms a basis for further
discussion, and shows that the patterns for current and voltage are similar.
(They can be related to the idea of exponential decay in radioactivity, Episode
513, if students have met this previously.)
Even though some specifications require the
use of data logging for this, it is worth collecting data manually from a slow
discharge and then getting the students to plot the graphs of current against
time and voltage against time for the decay.
The specifications do not require details of
the charging process but data for this is easily collected in the same
experiment.
Discussion:
Characteristics of exponentials
Draw out the essential features of the
discharge graphs. Sketch three graphs, for Q,
I and V against t. All start at
a point on the y-axis, and are
asymptotic on the t-axis. All have
the same general shape. How are they related?
The Q graph is simply the V graph multiplied
by C (since Q = CV).
The I graph is the V graph divided by R
(since I = V/R).
The I graph is also the gradient of the Q
graph (since I = dQ/dt).
Add small tangents along the Q graph to show this latter pattern. A
large charge stored means that there is a large pd across the capacitor; this
makes a large current flow, so the charge decreases rapidly. When the charge is
smaller, the pd must be lower and so a smaller current flows. (Students should
see that this will result in quantities which get gradually smaller and
smaller, but which never reach zero.)
Student activity:
Spreadsheet model
 |
Students can use an iterative approach, with the help of a spreadsheet, to see the pattern of potential difference across the capacitor while it is discharging (top graph), and charging (bottom graph).
Student experiment:
Varying R
and C
The previous experiment produced graphs of
the discharge for a particular combination of resistor and capacitor. This can
be extended by looking at the decay for a range of values of C and R. If a
datalogger is available, this can be done quickly and can include some rapid
decays. If a datalogger is not available, measurements can be taken with the
apparatus used earlier in TAP 129-1.
Before the experiment, ask your students how
the graphs would be affected if the value of R was increased (for a particular
value of pd the current will be less, and the decays will be slower); and if
the value of C was increased (more charge stored for a given pd; the initial
current will be the same, but the decay will be slower, because it will take
longer for the greater quantity of charge to flow away.)
Discussion:
Deriving exponential equations
At this point, you have a choice:
You can jump directly to the exponential
equations and show that they produce the correct graphs.
Alternatively, you can work through the
derivation of the equations, starting from the underlying physics.
We will follow the second approach.
To explain the pattern seen in the previous
experiment you will have to lead your pupils carefully through an argument
which will call on ideas about capacitors and about electrical circuits.
Consider the circuit shown:
 |
When the switch is in position A, the
capacitor C gains a charge Qo so that the pd across the capacitor Vo
equals the battery emf
When the switch is moved to position B, the
discharge process begins. Suppose that at a time t, the charge has fallen to Q,
the pd is V and there is a current I flowing as shown. At this moment:
I
= V/R Equation
1
In a short time Dt, a charge equal to DQ flows from one
plate to the other so:
I = –DQ / Dt Equation 2
[with the minus sign showing that the charge
on the capacitor has become smaller]
For the capacitor: V = Q/C Equation
3
Eliminating I and V leads to DQ = – (Q/CR) ´ Dt Equation 4
Equation 4 is a recipe for describing how any
capacitor will discharge based on the simple physics of Equations 1 – 3. As in
the activity above, it can be used in a spreadsheet to calculate how the
charge, pd and current change during the capacitor discharge.
Equation 4 can be re-arranged as
DQ/Q = – (1/CR) ´ Dt
showing the constant ratio property
characteristic of an exponential change (i.e. equal intervals of time give
equal fractional changes in charge).
We can write Equation 4 as a differential
equation: dQ/dt = – (1/CR) ´ Q
Solving this gives: Q = Qo e-t/CR where Qo = CVo
Current and voltage follow the same pattern.
From Equations 2 and 3 it follows that
I = Io e-t/CR where Io = Vo / R
and V = Vo e-t/CR
Worked example:
Using the equations
A 200 mF capacitor is charged to
10 V and then discharged through a 250 kW resistor. Calculate the pd
across the capacitor at intervals of 10 s.
(The values here have been chosen to give a
time constant of 50 s.)
First calculate CR = 50 s.
Draw up a table and help students to complete
it. (Some students will need help with using the ex function on
their calculators.) They can then draw a V-t graph.
|
t / s
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0
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10
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20
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30
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40
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50
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60
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etc
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V / V
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10
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8.2
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6.7
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5.5
|
etc
|
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I / mA
|
40
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Q / mC
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2000
|
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Explain how to calculate Io = Vo
/ R and Qo = CVo, so that they can complete the last two
rows in the table.
It is useful to draw a Q-t graph and deduce
the gradient at various points. These values can then be compared with the
corresponding instantaneous current values.
Similarly, the area under the I-t graph can
be found (by counting squares) and compared with the values of charge Q.
Discussion:
Time constant
For radioactive decay, the half life is a
useful concept. A quantity known as the ‘time constant’ is commonly used in a
similar way when dealing with capacitor discharge.
Consider Q
= Qo e-t/CR
When t = CR, we have Q = Qo
/ e-1
i.e. this is the time when the charge has
fallen to 1/e = 0.37 (about 1/3) of its initial value. CR is known as the time
constant – the larger it is, the longer the capacitor will take to discharge.
The units of the time constant are ‘seconds’.
Why? (F ´ W = C V-1 ´ V A-1 = C A-1 =
C C-1 s = s)
(Your specifications may require the
relationship between the time constant and the ‘halving time’ T1/2: T1/2 = ln 2 ´ CR = 0.69 ´ CR)
Student activity:
Analysing graphs
Students should look through their experimental
results and determine the time constant from a discharge graph. They should
check whether the experimental value is equal to the calculated value RC. Why
might it not be? (Because the manufacturer’s values of R and C, are only given
to a specified range, or tolerance, and that range is rather large for C.)
Student questions:
Practice with equations
Questions on capacitor discharge and the time
constant, including a further opportunity to model the discharge process using
a spreadsheet.
Discussion:
Back to reality
After a lot of maths, there is a danger that
students will lose sight of the fact that capacitors are common components with
a wide range of uses.
Some of these can now be explained more
thoroughly than in the initial introduction. Ask students to consider whether
large or small values of C and R are appropriate in each case. Some examples
are:
Back-up power supplies in computers, watches
etc., where a relatively large capacitor (often > 1F) charged to a low
voltage may be used.
Some physics experiments need very high
currents delivered for a very short time (e.g. inertial fusion). A bank of
capacitors can be charged over a period of time but discharged in a fraction of
a second when required.
Similarly, the rapid release of energy needed
for a ‘flash bulb’ in a camera often involves capacitor discharge. Try
dismantling a disposable camera to see the capacitor.
TAP 129- 1: Slow charge and discharge
This activity is intended to give students a
‘slow motion’ view of charging and discharging a capacitor and hence to show
the shape of the charge and discharge curves.
Requirements
1000 mF 25 V electrolytic capacitor
100 kW 0.6W resistor
clip component holders (2) or crocodile clips
(4)
9 V battery or very smooth dc supply
voltmeter f.s.d. 10 V digital
ammeter f.s.d 100 mA (2)
leads
Set-up:
Make the circuit shown below
Touch point A with flying lead, F, to make
sure that any charge that the capacitor was holding is released, thus
discharging the capacitor.
Make sure you can see the voltmeter and
ammeters clearly.
Have a table ready to record current and
voltage values every 15 seconds.
Connect flying lead, F, to 1. This will begin
charging the capacitor through the 100 kW resistor.
Carefully monitor what is happening on the
meters and note down values every 15 s for around 500 s.
Now connect the flying lead to 2 and, in a
similar way, monitor carefully what is happening.
Analysing your results
Sketch or plot graphs of how the following
vary as the capacitor charges and discharges:
the current flowing into a capacitor;
the voltage across a capacitor.
Note any similarities and differences between
pairs of graphs or all the graphs.
Try to explain the shapes by thinking about
electron flow into and out of the capacitor. It may help you to recall a basic
rule of electrostatics: ‘like charges repel’.
Use your graphs of the discharge to determine
the time for the voltage to fall to 1/2, 1/4 and 1/8 its original
value.
Is there a pattern?
External references
This activity is taken from Salters Horners
Advanced Physics, A2, Section Transport on Track, TRA, Activity 20
TAP 129- 2: One step at a time
Use a spreadsheet to examine the discharge of
a capacitor over a succession of small time intervals.
Introduction
A simple model of capacitor discharge needs
to follow a loop of operations to carry out the following steps:
Set the initial values for R, C, and V. Decide
on the interval of time you are going to use, Dt, and
input that. Label your columns.
Calculate an initial charge Q stored on the capacitor using Q = CV
Find the initial current, I, in the circuit using I = V / R where
again R needs to be set
initially.
Work out the small change in charge DQ that
has occurred in a small time interval Dt (which the
computer will need to be told) due to the current in the circuit. The equation
is
DQ = −I × Dt and the
negative sign implies that discharge is taking place, as opposed to charging.
Calculate the new charge Q ready for the next cycle of the
loop using Q = Q + DQ
Calculate a new value for V using V = Q / C
Calculate elapsed time, t, ready for the next loop using t = t + Dt
(NB. The spreadsheet must be programmed to
set t = 0 initially.)
Return to step 2 and repeat steps 2 to 8.
Using the spreadsheet
For your first try at this spreadsheet, try
the following set-up:
series resistance R = 100 000 W
capacitance C = 0.000 05 F
supply voltage V = 10 V
time interval Dt = 0.5 s
initial charge from Q = CV = 0.0005 C
|
Time t/s
|
Charge Q/C
|
Current ImA
|
Change in charge
DQ/C
|
Voltage V/V
|
|
0.0
|
0.000 500
|
0.000 100
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−0.000 050 00
|
10.00
|
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etc.
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Try using the spreadsheet to do the
following:
Generate a graph of how voltage V across the
capacitor varies with time.
By changing the initial values of R and C,
observe how the rate of decay varies. Does it fit the equation for exponential
decay, V = Voe−t/RC?
Compare the discharge with the calculated
value of time constant (RC). Show by rearranging the equation above that V = Vo/e
when t = RC. Then confirm this from your spreadsheet.
Further work
The formula for capacitor charging is V = Vo
(1 − e−t/RC) = Vo − Voe-t/RC
Adapt your spreadsheet to model a charging
capacitor by just adding an extra column based on the formula above. Try this
for at least one of your original RC combinations. You may wish to display only
the time and final voltage columns this time.
Practical advice
This spreadsheet activity has several
purposes. One is to help students to think about what is actually happening
when a capacitor discharges. Another is to reinforce the use of Q = CV. Another
is to give students some experience of iterative numerical modelling. For
capacitor discharge, we can derive analytical expressions for the way the
variable of interest changes with time, but in other situations an analytical
approach is not always easy or even possible, so tracing a change over a series
of small time steps is sometimes the only sensible way to predict what will
happen.
Students may need help in setting up the
spreadsheet.
External references
This activity is taken from Salters Horners
Advanced Physics, A2, Section Transport on Track, TRA, Activity 23
Looking forward
Use automated data capture to study the decay
of potential difference across the terminals of a capacitor as it discharges
through a resistor.
The exact procedure will have to be modified
depending on what software is available with your particular data capture and
analysis package. If you are handing out these instructions to pupils you will
need to edit these instructions accordingly.
You should aim to process the results by some
computer based method.
The option quoted here is to identify the
mathematical form of the potential difference against time graph using a
curve-fitting method, and to then evaluate the decay constant.
Alternatively, if you have covered log
graphs, you could use a log plot to identify the decay constant.
You will need
computer, running data capture and analysis software
data capture and analysis software
data capture device
voltage sensor
capacitor 100 mF
clip component holder
resistance substitution box
spst switch
power supply, 5 V dc
leads, 4 mm
Getting some data
Connect the capacitor C and resistor R
in parallel as shown in the diagram. Add the switch and battery to the circuit
as shown and attach the leads from the voltage sensor across the resistor.
Check that the polarity of the capacitor terminals matches the + and –
connections to the battery.
Prepare the computer program to record
potential difference for 10 seconds. Conduct a trial run of the experiment as
follows:
· set the program to start recording data
· close the switch to charge the capacitor
· open the switch.
As the capacitor discharges through the
resistor, you can observe the graph of the decaying potential difference across
the capacitor and resistor. If necessary, adjust the scale of the potential
difference axis to give a large clear display of the graph.
You may wish to adjust the start condition
for logging so that logging only begins when the potential difference falls
below a chosen value. This is called a ‘triggered’ start. To identify a
suitable value for the trigger potential difference, choose a value which is
slightly less than the maximum potential difference shown on the trial graph.
Clear the previously collected data from the program
set it ready to start logging and perform the experiment as you did in
the trial run.
If your data analysis software has this
function, select a trial fit. Choose the formula y = a e bx + c and adjust the buttons to fit and plot the data appropriately.
You may find that a new curve is plotted directly over the data collected in
the experiment. Make a note of the value of ‘b’ in the formula. This is called the decay constant of
the curve. Notice that it is negative and record the value in the table.
|
Capacitance / mF
|
Resistance / W
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RC / F ´ W
|
1 / RC
/ s–1
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Decay constant
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100
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10 000
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1.0
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etc
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Repeat the experiment with different values
of C and R, overlaying each new graph on the
previous one to build a set of graphs.
Analysing the results
What is the connection between the decay
constant and the shape of each graph?
The decay constant is a measure of how
quickly the potential difference falls to zero. Look at the values calculated
by the program to confirm that this is true.
Theory predicts that the decay constant is
given by 1 / RC. Do your
results provide evidence for this? (The value of RC is known as the time constant of the circuit. Notice that,
being the inverse of the decay constant, it is an indicator of how long
it takes the voltage to decay rather than the rate of decay.)
Thinking about the theory
The potential difference across the resistor
is given by V = I R.
Now the current I is due to the discharging capacitor where I = – dQ / dt.
For the capacitor Q = C
V and dQ = C dV. So we can say
or
Since R and C are constant in
each experiment, this predicts that dV / dt is proportional to V. This can be put to the test
by plotting a graph of V
against dV / dt and looking for a straight line
with a gradient of ‘– RC ’.
(If you are using Insight, you can test this
as follows.
Select ‘Define’ from the ‘Edit’ menu. If the
potential difference data are stored in channel A, build a formula of the form
‘dA / dt’ and store the calculated data in channel F. Adjust the axes
to plot A versus F. This is equivalent to V against dV / dt. Select ‘Gradient’ from the ‘Analyse’ menu to find the gradient of the
line. How does this compare with the value of – RC?)
You have learned
1. The graph of the potential difference
across a discharging capacitor is an exponential function.
2. The decay constant of an exponential
graph is a measure of how quickly the graph falls to zero.
3. The
decay constant for a discharging capacitor is given by 1 / RC.
4. The rate of decay in the potential
difference (dV / dt) varies in proportion to the potential
difference (V).
Practical advice
This activity gives experience in using
software for managing the collection of data, plotting graphs, analysing the
data and calculating new data.
The purpose of the practice run is to allow
the student to gain confidence in handling the equipment and software together
and to enable necessary adjustments to be made to the axis scales.
Use the software to display the graphs of
several experiments on the same axes. It is desirable to employ the ‘triggered
start’ facility so that all the graphs obtained start from the same initial potential
difference at time zero, or subsequently by software manipulation.
The curve-fitting method described here gives
a novel way of evaluating the decay constant which emerges as one of the
parameters calculated by the program in the fitting process. Students are
encouraged to make a link between the decay constant and the shape of the
curve; a quick discharge is associated with a relatively large decay constant
and vice versa. The further link between the decay constant and the time
constant (RC) is pointed out.
If the software allows it, (a “Ratio” or
“Analyse” menu perhaps) measure the time constant directly as the time for the
potential difference to fall to 1 / e (37%) of an initial value. This can then
be compared with the value of RC.
The further analysis of the data by
calculating dV / dt and plotting a new graph might be regarded as an optional
extension. This method follows the traditional analysis which seeks linearity
and calculates RC from the gradient. Students can also be asked to use a ‘Trial
fit’ to obtain a best fit straight line for the data. The program calculates
the gradient as a parameter in the straight-line formula.
Alternative approaches
The same procedure may be used with a
conventional voltmeter instead of the computer and interface. The data could be
entered into a spreadsheet and manipulated to calculate speeds and plot graphs.
However, the method described here considerably reduces errors due to taking
readings in quick succession and removes the trouble of entering much of the
data.
Be safe
Care should be taken to ensure that
electrolytic capacitors are connected with the correct polarity and that the working
voltage is not exceeded.
External
references
This activity is adapted from Advancing
Physics Chapter 10, 130E
A 250 mF capacitor is charged through a 100 kW resistor.
1. Calculate
the time constant of the circuit.
2. The
initial current is 100 mA. What is the current
after 30 s?
3. Suggest values of R and C
which would produce RC circuits with time constants of 1.0 s and
20 s.
4. The insulation between the plates of
some capacitors is not perfect, and allows a leakage current to flow, which
discharges the capacitor. The capacitor is thus said to have a leakage
resistance. A 10 mF capacitor is charged to a potential
difference of 20 V and then isolated. If its leakage resistance is 10 MW, how long will it take for the charge to fall to 100 mC?
A 100 mF capacitor is charged and connected to a digital voltmeter (which
has a very high resistance). The pd measured across the capacitor falls to half
its initial value in 600 s.
5. Calculate
the time constant of the discharge process.
6. Calculate
the effective resistance of the capacitor insulation.
Practical advice
These questions require familiarity with time
constants and the exponential function.
Answers and worked solutions
1. 25
s
2. 30
mA
3. For
example, C = 10 mF and R = 100 kW; for example, C = 200 mF and R = 100 kW.
4. 69
s
5. 8.6
× 102 s
6. 8.6
MW
Worked solutions
1.
R C = 250 ´ 10–6 F ´ 100 ´ 103 W = 25 s
2.
3.
Any pair of values such that R C = 1.0
s and R C = 20 s is acceptable.
4.
.Since
and
5.
so
and
6.
External references
This activity is taken from Advancing Physics
Chapter 10, 70S
A capacitor is charged to a potential
difference of 1.0 V. The potential difference is measured at 10 s intervals, as
shown in the table. When t = 15 s, a resistance of 1.0 MW is connected across the capacitor terminals.
|
t / s
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V / V
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0
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1.00
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10
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1.00
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20
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0.81
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30
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0.54
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40
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0.35
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50
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0.23
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60
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0.15
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1. What
is the current in the resistor at t = 15 s?
2. Plot a graph of V against t,
and measure the rate of decrease of V immediately after
t = 15 s.
3. Using
the relationship I = d Q / d t
= C d V / d t, calculate the value of the capacitance.
4. Plot also a graph of ln V
against t to show the exponential decay of voltage, and use the gradient
to find the time constant (RC).
5. From
the time constant calculate the capacitance C.
6. Explain
which method gives a better value of C, and why.
Practical advice
This question could usefully follow an
experiment where students see capacitors discharge. It also shows that
exponential behaviour can be demonstrated by the use of natural logarithms.
Answers and worked solutions
1. 1
mA
2. Approximately
3.8 × 10–2 V s–1
3. Approximately
26 mF
4. 24.5
s
5. 24
mF
6. The second method is better, because it
avoids the difficulty of accurately drawing a gradient for the V–t graph
at 15 s.
Worked solutions
1.
2.
3.
4.
|
t / s
|
ln V
|
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0
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0
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10
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10
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20
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–0.21
|
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30
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–0.62
|
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40
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–1.05
|
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50
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–1.47
|
|
60
|
–1.90
|
5.
6. The second method is better, because it
avoids the difficulty of accurately drawing a gradient for the V–t graph at 15
s.
External references
This activity is taken from Advancing Physics
Chapter 10, 80S
Tap 129- 6: Capacitors with the exponential
equation
You will find it useful to be able to use the
equation to calculate, for example, the pd across a capacitor after it has been
discharging for some time. The following question give you practice in doing
this.
A 10 F capacitor is charged to 5.0 V and then
discharged through a 5 kW resistor.
1. Calculate
the time constant for the circuit.
2. How
much energy is stored in the capacitor when it is fully charged?
3. Calculate
how long it takes for the pd across the capacitor to fall to 4.0 V.
4. How
much energy will have been transferred from the capacitor during this process?
5. What
will be the pd across the capacitor after 5.0 ´ 104 s?
6. Calculate the time taken for 50% of the
capacitor’s initial energy to go into heating in the resistor.
Practical advice
These questions involving solving problems
using the equation may only be suitable for those aiming at higher grades, or
the mathematically more adept.
Answers and worked solutions
1. 5.0
× 104 s
2. 125
J
3. About
3 hours
4. 45
J
5. 1.8(4)
V
6. 1.7
× 104 s
Worked solutions
1.
2.
3.
so
gives
4.
Energy at 4.0 V
so the energy lost is 45 J.
so the energy lost is 45 J.
5.
6. For 50% energy:
so
which gives t = 1.73 x 104 s,
or just under 5 hours.
so
which gives t = 1.73 x 104 s,
or just under 5 hours.
External references
This activity is taken from Advancing Physics
Chapter 10, 140S
Introduction
One use of capacitors is as backup sources of
energy in the case of power failure in, for example, memory systems, where
total loss of power may result in the loss of essential data. Such capacitors
operate at low voltages but generally have very high capacitance. These
questions illustrate how knowledge of the mathematical model which describes
the exponential decay of the charge on a capacitor is put to good use in
choosing a suitable capacitor for a given application.
These questions involve working with data on
such commercially available capacitors and using an Excel spreadsheet to
examine the properties of a range of such devices.
What to do
There are data tables included giving details
of high-value capacitors. You will be asked questions about the use of these
and also asked to open up a spreadsheet and work with it. Work through the
questions in order and follow the instructions given to help you.
High capacitance memory back-up capacitors
A range of high capacitance memory back-up
capacitors for applications which require memory retention under power failure
conditions. Such applications include automotive energy management systems,
domestic and industrial control systems.
|
Technical specification
|
|
|
Operating temperature range
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–40°C to +70°C
|
|
Capacitance tolerance
|
–20% to +80%
|
|
Rated voltage
|
2.5 V
|
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Value / F
|
Inflow I / mA
|
Body
|
Lead
|
||||||
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|
|
L
|
dia
|
L
|
dia
|
pitch
|
|||
|
0.33
|
0.12
|
21
|
6.8
|
15
|
0.7
|
2.5
|
|||
|
1.0
|
0.18
|
22
|
8
|
15
|
0.7
|
3.5
|
|||
|
3.3
|
0.30
|
23
|
12.5
|
15
|
0.8
|
5.0
|
|||
|
10
|
0.6
|
35
|
18
|
15
|
0.8
|
7.5
|
|||
|
Value
|
price per item
|
||
|
F
|
1–9
|
10–49
|
50–99
|
|
0.33
|
£1.90
|
£1.75
|
£1.62
|
|
1.0
|
£2.09
|
£1.93
|
£1.78
|
|
3.3
|
£2.62
|
£2.41
|
£2.23
|
|
10
|
£4.39
|
£4.05
|
£3.73
|
A customer requires a capacitor as a backup
for a power supply that can provide a voltage which does not drop below 2.0 V
for several hours. The effective load (resistance) connected across the power
supply is 5 W. You are required to choose a capacitor from the selection above.
This spreadsheet allows you to examine the discharge curve of capacitors that
you select.
Open
the Excel Spreadsheet see TAP
129-8
When you open up the spreadsheet you will see
that a capacitor of 10.0 F has been chosen connected to a load of 10 kW. The charge and pd across the capacitor are calculated every half
hour and displayed in the table and on the named graphs. You can change the
initial conditions like this:
1. To
change the time interval to an hour, type 1.0, press the arrow key or enter.
2. Then press key F9.
The spreadsheet will not be recalculated until you press F9.
The spreadsheet will not be recalculated until you press F9.
3. Try changing the value of capacitance
and look at the effect on the table and on the graphs.
You can change other quantities in the same
way, for example, the load resistance and the time interval. Remember that the
load must be in kW and the time in hours.
When you are satisfied that you can handle
the spreadsheet confidently, switch back to the questions.
Questions
1. Calculate the amount of energy which
can be stored in the 10 F capacitor when charged to 2.5 V.
2. The capacitor in question 1 is
connected across a 10 kW resistor. What is the time
constant of this circuit?
3. What
is the significance of the time constant?
Back-up power alkaline battery packs
A 4.5 volt alkaline battery used to back-up
memory and clocks in laptop and desktop computers.
|
Technical specification
|
|
|
Voltage
|
4.5 V
|
|
Capacity
|
1200 mAh
|
|
Temperature range
|
~30 °C to 60 °C
|
|
Dimensions
|
58 × 23 × 26 mm
|
|
price per item
|
|
|
1–9
|
10–24
|
|
£10.40
|
£9.36
|
4. Look at the second line in the above
table called the ‘capacity’. Note that this is the product of current and time.
Convert the value given to coulombs.
5. Use the data to estimate how much
energy this battery stores. (Hint: look up the definition of potential
difference if necessary.)
6. Compare the energy available from this
battery with that stored in the capacitor in question 1.
To answer the next
question, use the spreadsheet.
7. Use the data in the table above and the
spreadsheet to determine which capacitor will provide a backup voltage of not
less than 1.5 V for at least 2 hours into a load of 10 kW. Assume that the second column gives the maximum current allowed.
8. Did you identify more than one suitable
capacitor from the list? If so, which would you choose and why?
9. What are the factors you would consider
if you had to decide between a battery or a high-value capacitor for backup
energy supply? What would be your choice in this case and why?
10. Another user needs a capacitor to
provide a pd across a load of 1.0 MW which does not
fall more than 10% from 2.5 V in 24 hours. Is any one of the above capacitors
suitable?
Taking it further – checking the spreadsheet
model
Go back to the spreadsheet. You will see that
there are two columns containing calculated values of charge, Q. These
have been done in two different ways. The column on the far right has been
calculated using the exponential equation
and the capacitor equation
This equation links the two continuous
variables Q and t and gives ‘exact’ values for Q. The
column on the far left has been calculated by an iterative method based on the
differential equation
in which each successive value of Q is
used to calculate the next one, knowing the current flowing for a short (but
finite) time. It is a useful exercise to compare the two methods by
superimposing the two graphs of Q against t. You will be able to
see that they differ. If you change the value of the time interval D t over which each calculation is done, you will see that the
amount of agreement between the two graphs also changes. Experiment with D t to see what you have to do to improve the agreement. How
far you get with this will depend on how practised you are with using
spreadsheets. You may find it useful to change the scale on the axes. To do so,
click on the axis for which you want to modify the scale and select ‘format’
then ‘selected axis’. You can rescale the maximum and minimum values of that
scale. Select the other scale and repeat if necessary.
Practical advice
This unit contains a range of activities
designed around the exponential decay of the charge on a capacitor. The use of
data from the current RS catalogue sets the work into a relevant modern
context. Before starting the unit, students should be familiar with the basic
capacitor equations Q = C V and W = ½ Q V etc. They
will also need to be able to use a spreadsheet and, although some guidance is
given, it may not be enough for students who are less confident. If students
can work together on a network, they could be taken through some of the steps
as a group exercise, or perhaps can be helped individually if time and staff
are available. The spreadsheet calculations are done by iteration and by using
the exponential equation, and the spreadsheet work later in the unit focuses on
these aspects, hopefully helping students to understand the idea that the
iterative calculation approaches the analytical solution as Dt → 0.
Social and human context
There is an element of economic awareness in
this work, in that students make a choice between capacitor or battery backup.
Answers and worked solutions
1. 31
J
2. 105
s
3. Time
for charge or pd to fall to 1 / e of its original value.
4. 4.3
× 103 C
5. 1.9
× 104 J
6. Much
larger
7. 3.3
F or 10 F
8. Yes, see above answer. Choose 3.3 F,
which is cheaper, or choose 10 F, which has more margin for error in
capacitance.
9. Consider size, cost, operating
temperature, lifetime; choose capacitor – cheaper, can be recharged, so longer
life likely.
10. Yes,
10 F will do (3.3 F will not).
External references
This activity is taken from Advancing Physics
Chapter 10, 130D
TAP 129- 8 Excel Spreadsheet for TAP 129-7
The
excel spread sheet contains the following data:
And the graphs obtained are:-
External references
This activity is taken from Advancing Physics
Chapter 10, 130D
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