8. THE CAPACITOR
You will study several aspects of a capacitor, how the voltage across it changes with time
as it is being charged and discharged and how it stores energy.
Theory
Why Study a Capacitor?
The most well known device for storing electrical
energy is no doubt the battery. But other devices
have the ability to store energy—the capacitor and
inductor to name two. Since the capacitor is the
easier of the two to study quantitatively, and the
easier to understand in terms of the physics, it is
the subject of study in this experiment.
But capacitors are worthy of study for other
reasons. Most consumers are aware that in many
of the technological devices they buy there is a
limitation placed on frequency—or how quickly in
the devices voltages and currents can change. The
speakers in our audio system might respond to a
maximum of 15 kHz, or the CPU in our computer
system might function at a maximum “speed” of
200 MHz. Serial or ethernet cables might not be
able to sustain a certain bandwidth beyond a
certain distance, and so on.
These facts are manifestations of capacitance
and inductance in the circuits. Capacitors and inductors have impedances (or effective resistances)
that depend on frequency; the effects therefore also
depend on frequency. This can be described as
“electrical inertia”, a reluctance for the circuits to
respond instantaneously to electrical changes. The
object of this experiment is to study one example
of this reluctance. For practical reasons we focus
not on AC changes, but rather on a DC current and
voltage that changes with time slowly enough to
be easily observed.1
The Idea of Electrical Inertia
Recall that in the experiment “DC Circuits”, you
measured voltage and current with digital multimeters some time after you connected up the
source of electrical energy (using a circuit as is
shown in Figure 1a). In fact, you may have found it
desireable to wait to enable the circuit to come to
an “electrical equilibrium”.2 In any case, this time
was so short that for all practical purposes equilibrium was instantaneous (with a change in voltage
as is shown in Figure 1b).
If it were possible to use a high speed device to
measure the voltage across the resistor during the
very short time just before and just after the connection, the voltage response might resemble
Figure 1c. The voltage cannot go instantaneously
from 0 to V volts, but rather, requires some time to
do so. This is an observation of electrical inertia.
This effect is caused by very small capacitive and
inductive components in what might otherwise be
regarded as an ideal resistor. In this experiment
this effect is deliberately emphasized for study by
using a capacitor instead of a resistor.
The Capacitor
A capacitor is a device with the ability to store
electrical energy (and therefore with a built-in
electrical inertia). It is commonly made from two
pieces of metal called plates separated by an insula-
ting material called a dielectric. A capacitor’s
capacitance depends on the size of the plates, the
distance between the plates and the ability of the
dielectric to respond to an electric field.3 It is found
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8 The Capacitor
experimentally that the voltage developed across a
capacitor is proportional to the charge q (in coulombs, abbreviated C) stored in the capacitor. That
is,
q = Cv ,
…[1]
where the constant of proportionality, C, is the
capacitance. C has the unit farad (abbreviated F). 1 F
= 1 C.V–1. If v in eq[1] is time dependent then q is
also time dependent.4 Though the capacitor used
in this experiment is nominally 1 farad, a farad is a
very large capacitance. More typical values are 0.1
µF (used at audio frequencies) and 0.01 µF (used at
radio frequencies). (µ means micro, or 10 –6.) The
value of C depends on the construction of the capacitor, especially on the conductors and the kind of
dielectric used. Details on the special construction
of the capacitor used in this experiment are given
in the Addendum. You can read this at your
convenience.
S
V=
5 volts
V
R
A
(a)
V
5
V
5
t0
time t
t0
(b)
time t
(c)
Figure 1. When the switch S is closed at time t 0 in a typical DC circuit shown in (a) the voltage across the resistor, if
observed on a multimeter, would be seen for all practical purposes to rise instantaneously from 0 to V volts (b).
However, if a high speed measuring device were used the voltage might be seen to rise in a curve as shown in (c).
The curve is a manifestation of electrical inertia. The time axis of (c) is expanded with respect to (b).
Charging/Discharging a Capacitor
Experimental Details
The capacitor is actually charged and discharged
by means of the circuit shown in Figure 2. Figure 2
has certain features in common with Figure 1a, but
the source of electrical energy here is a power
supply not a battery. The capacitor is charged by
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connecting it to the power supply (switch S1 is
thrown to the left). Subsequently the capacitor is
discharged by connecting it to the load resistor R
(switch S1 is thrown to the right). The charge
stored in the capacitor flows through the resistor
The Capacitor 8
producing a voltage drop across the resistor. This
voltage could be measured with a digital multimeter (as described above), but is here measured
by means of a Serial Box Interface (SBI) and a
computer. The use of an SBI avoids the problems
associated with relatively slow responding digital
devices like multimeters. And the computer also
plots a graph of voltage versus time automatically
(not to mention making possible subsequent data
analysis).
S1
S2
red
lamp
30 Ω
to power supply
1.0 F
30 Ω
black
to SBI
Figure 2. Schematic of the circuit box used in this experiment (within the dashed outline) which you can see through
its transparent cover. Cables on the left plug into the power supply. Cables on the bottom coming from across the
capacitor go to the SBI. Switch S1 controls charge/discharge, switch S2 provides a way of shorting one of the 30 Ω
load resistors. The purpose of the lamp is to provide a visual indication of the current flowing through the capacitor
as it is being charged.
Charging/Discharging a Capacitor
Mathematics
Assumption
In what follows we assume that when charging,
the capacitor begins from a fully discharged state
and when discharging begins from a fully charged
state.
moved from the source to the capacitor in increments d q. For each increment the voltage is v. The
element of work done d w by the supply in transferring a charge dq to the capacitor when at voltage
v is given by
Energy Required to Charge
During charging the capacitor is brought from a
“begin” state when the voltage across it is zero to
an “end” state when the voltage is V volts. The q
and v in eq[1] change with time (which we examine in the next section), but eventually the voltage
reaches some maximum V volts and the charge a
maximum Q coulombs.
A little calculus is useful here. Charge is
dw = vdq
= Cvdv ,
using eq[1]. Thus the total work W done is given
by the integral
W=
∫
V
0
V
dw = ∫0 Cvdv
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8 The Capacitor
=
1
CV 2 ,
2
…[2a]
=
1
QV ,
2
…[2b]
1 Q2
2 C
…[2c]
=
vC (t) – Ri(t) = 0 ,
joules, by successive substitutions of eq[1]. The
energy transferred to the capacitor can be calculated from any of these equations.
Voltage Change During Charging
We now examine the physics as v is changing.
Suppose at time t = 0, S1 is moved from the right to
the left (Figure 2) to begin the charging process. By
applying Kirchoff’s voltage rule around the loop at
some later clock time t when the capacitor’s charge
is q(t) and current i(t) is flowing we have
V–
q(t)
– R'i(t) = 0 ,
C
where R’ stands for the effective resistance in the
charging circuit comprising lamp and power supply (not to be confused with the load resistor R in
the discharging circuit). Substituting i = dq/dt and
eq[1] and rearranging we obtain an equation in
v(t):
dv(t) v(t)
V
+ ' = ' .
dt
RC RC
…[3]
This equation is a first order differential equation
with solution v(t). You should be able to show that
a solution is5
t
–


RC
v(t) = V 1– e  .


…[4]
The quantity R’C (with units of s) is called the time
constant. This is the time required for the voltage to
rise to within 1/e of V. A complementary treatment applies for the discharge of the capacitor.
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Voltage Change During Discharge
We suppose that at some clock time t = 0 seconds,
S1 is thrown to the right (Figure 2) to begin the
discharge process. By applying Kirchoff’s voltage
rule around the loop at some clock time t when the
voltage across the capacitor is vC(t) and a current
i(t) is flowing we have
…[5]
where now R is the resistance of the load resistor.
Using eq[1] and the fact that
i(t) = –
dq(t)
,
dt
(as current flows charge decreases) it can be shown
that
dv(t) v(t)
+
=0.
dt
RC
…[6]
This equation is a first order differential equation
in v(t). You should be able to show that a solution
is
–
v(t) = Ve
= Ve
where
t
RC
–
,
…[7]
t
tC
tC = RC .
…[8]
Eq[7] is a relatively simple exponential function as
is seen in Figure 3. tC is called the time constant and
quantifies how sharply the voltage decreases. If R
is in units of ohms and C in units of farads then RC
has the unit of time in seconds. tC is the time in seconds for the voltage to fall from the initial voltage
V to V/e = 0.368 V. The bigger the time constant
the less is the rate of voltage decay. Objectives of
this experiment are to test the validity of eqs[4]
and [7] during charge and discharge and to examine the corresponding changes in energy.
The Capacitor 8
vC (t)
V
V = .368 V
e
tC = R C
time t (s)
Figure 3. The exponential decay of the voltage across a capacitor discharging through a resistor.
The Experiment
Exercise 0. Preparation
Orientation
Many of the technical aspects of this experiment
you have seen before in other experiments.
LoggerPro is used here too. You will use a Serial
Box Interface (SBI) instead of a ULI board to detect
the voltage across the capacitor but the functioning
is similar.
with the guidance of Figure 2.
3 Confirm that the switch S1 is in the discharge
position.
3 Confirm that the switch S2 selects only one of
the load resistors.
3
Ô Identify the apparatus from Figure 4. You have
a source of electrical energy (MW regulated
12V power supply), a circuit box (containing
capacitor, switches and load resistors) connected to the output of the supply, lines running
from the circuit box to the Serial Box Interface
(SBI), and in turn to the computer.
3 Confirm that the power supply is OFF.
3 Using what you have learned in the experiment “DC Circuits” note the resistance of the
two load resistors and their uncertainties.
About the Circuit Box and the Capacitor
For convenience you may wish to orient the circuit
box so that it appears as in Figure 2 with switches
S1 and S2 arrayed along the top.
Ô The capacitor is the green cylindrical object
mounted upside down inside the circuit box.
3 Follow what you can see of the circuit diagram
through the transparent cover of the box and
3
¬
Á
Â
Ã
Setting up the SBI
Confirm that the power adapter for the SBI is
plugged into the power bar and the SBI is ON,
i.e., the light on the SBI is glowing green.
Confirm that the lines from the circuit box are
plugged into Port 1 of the SBI, and that the signal output cable from the SBI is plugged into
the printer port of the computer.
Opening LoggerPro
Boot the computer as you learned to do in the
experiment "Linear Motion".
Log into your account on the college network.
Remember, if you can’t log in you can always
save your experiment in the “Student Temp
Save” folder on the local hard drive. You can
log in or out at any time.
On the local hard drive "Macintosh HD" locate
and open folders in this order: “Physics” >>
“PHYA10/A20”>>“8. The Capacitor”.
Inside “8. The Capacitor” double click the icon
“The Capacitor Alias”. LoggerPro and The
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8 The Capacitor
Capacitor setup should run.
The Opening Screen
Examine this screen and identify these aspects:
Ô The calibration is set to display voltage between 0 and 5 volts.
Ô The program once started will log data continuously for 2 minutes. At the end of that time
logging will stop automatically.
Ô The calibration is set to collect two voltage
measurements per second making for 240
datapoints.
Ô The voltage that is being measured at the
present time is displayed at the bottom center
of the screen. This is the voltage that is logged
when you click the Collect button.
S1
MW 12V
power supply
Entering Formulas
LoggerPro logs voltage; it must be deliberately
instructed to calculate power. You have to enter
the formula into LoggerPro’s spreadsheet as you
did in other experiments. If you have forgotten
how to do this select Data >> New Column >>
Formula. Figure 5 shows a fragment of the screen
which then appears. Enter the formula for power
(“potential”^2/R) including the load resistance
you have selected via S2.
Changing the Graph Displayed in an Area
Recall that you can change the graph that is displayed in a graph area. To do this place the pointer
over the graph label (down the left hand side of the
graph), click the mouse and from the dialog box
select the graph you wish displayed. Do this now
to display the “Potential (Volts)” graph.
S2
circuit box
printer port
SBI
Figure 4. The equipment used in this experiment. The circuit box is shown in detail in Figure 3.
Figure 5. A fragment of the screen that appears when New Column is chosen.
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The Capacitor 8
Exercise 1. Collecting the Data and Fitting a Function
The objective of this exercise is to collect data, obtain graphs for charging and discharging and to fit the graphs to obtain the time constants.
Charging
You may wish to begin with S2 in the position to
short out one of the 30Ω load resistors. Turn the
power supply ON, throw S1 to the charging
position and click Collect. As the charging
proceeds observe the brightness of the lamp and
the voltage across the capacitor as printed at the
bottom center of the screen. Here are some
questions of a qualitative nature:
Questions
? Is the charging instantaneous?
? Does the charge curve at least look exponential?
? Roughly how long does it take for the voltage
to reach maximum?
? How bright is the lamp when the maximum
voltage is reached? Why?
?
Explain what is happening as the lamp dims
and then goes out.
Saving the Data
At the end of the 2 minutes you can save the data
you have just collected. Select Data >> Save
Latest Run.
Discharging
Now throw the switch to the discharge position
and click Collect. After 2 minutes you should see
a graph resembling Figure 6.
Questions:
? Does the capacitor completely discharge in 120
seconds?
? Does the capacitor charge faster than it discharges? What is a likely explanation for this?
Graph produced by LoggerPro, copied to the clipboard and pasted directly into Microsoft Word.
Figure 6. An example of charge and discharge curves placed on the same graph. The little information windows can
be placed anywhere on the graph by dragging with the mouse.
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8 The Capacitor
Finding the Time Constants
You can conclude that the discharge is exponential
only if the voltage function given by eq[7] really
does describe your data. And if it does describe
your data then the results of the fit should yield
the time constant tC. Accordingly, apply what you
have learned about fitting from other experiments
to fit an exponential function to the decay curve.
What value for the time constant does the fit
return?
Questions:
? Can you estimate this value’s uncertainty?
? How well does this “experimental” value
agree with the “theoretical” value predicted by
eq[8]?
? Assuming that R has the unit ohms and C has
the unit farads, prove that tC has the unit
seconds.
Exercise 2. Aspects of Energy
Energy In
Based on eq[2a] how much energy did the power
supply deliver when charging the capacitor to the
maximum voltage V? Did the power supply
actually deliver more energy than this? Explain.
Energy Out
Display the power vs clock time graph. How is the
energy stored in the capacitor related to the power
graph? By diffentiation? Integration?
3 Find the energy stored from the power vs clock
time graph. For guidance an example is shown
in Figure 7. How well does this number agree
with the number obtained in the first
paragraph of this exercise?
Graph produced by LoggerPro, copied to the clipboard and pasted directly into Microsoft Word.
Figure 7. A typical output from LoggerPro showing the area under a power vs clock time curve. The area, given as
12.1 J differs by only about 3% from the expected value of 12.5 J.
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The Capacitor 8
Exercise 3. Repeat
Repeat the above with both load resistors in the circuit.
Questions:
?If R is increased by a factor of 2 what in theory is the effect on tC?
? Does the answer to the previous question mean that in this particular experiment a discharging
capacitor “loses” charge “faster” or “slower” than a charging one?
Physics Demonstrations on LaserDisc
from Chapter 46 Capacitance and RC Circuits
Demo 18-19 Parallel Plate Capacitor
Demo 18-28 RC Charging Curve
Activities Using Maple
E08The Capacitor
If you choose you can input the data you collected in the experiment “The Capacitor” into this worksheet,
plot it and fit it. This worksheet also attempts to shed light on the non-linear aspects of the voltage decay
you have no doubt observed. As a tutorial, the differential equation describing the decay of the voltage
across the capacitor is solved under assumptions of linear and non-linear decay.
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8 The Capacitor
Addendum. The Supercap 1 Farad Capacitor6
The capacitor used in this experiment is the NEC Supercap 1 farad, developed to protect memory
circuits during temporary power losses. The two materials serving as interface media in this capacitor are
activated carbon and sulfuric acid. Activated carbon is ground to a powder and mixed with sulfuric acid
to form a paste. The surface area of the activated carbon is approximately 1000 m2 per gram. The
combination of this very large surface area with the very small distance between the capacitor plates
results in a capacitance of about 200 to 400 farads per gram of activated carbon.
One problem with these materials is that if more than 1.2 volts is applied to this junction, the
aqueous sulfuric acid solution breaks down. To remedy this, the Supercap incorporates a number of small
cells, each consisting of a “sandwich” of the paste separated by a porous material. Each cell is
encompassed by an impermeable gasket (to hold everything together) squeezed between two electrically
conductive ends. The whole system is pressurized. Individual cells are stacked in series (end to end) and
squeezed into a metal outer cover. By stacking them in series, the effective voltage of each cell is
multiplied to useful proportions.
Stuart Quick 97
EndNotes for The Capacitor
1
Here we regard a DC current as one that flows in only one direction through a circuit; its amplitude however
may change with time.
2
Some waiting was required more due to the fact that the multimeters were digital devices and responded to
changes relatively slowly.
3
This is quantified by the dielectric constant. All things being equal a capacitor’s capacitance can be increased by
choosing a dielectric with a larger dielectric constant.
4
Lower case letters are used in eq[1] to indicate quantities which may be variables. Eq[1] applies whether or not q
and v are time dependent. When the voltage is a maximum then so also is the charge, ie., Q = CV.
5
It is sufficient to show that eq[4] satisfies eq[3] by direct substitution. Take the first derivative of eq[4] and
substitute it into eq[3] to show that the LHS = RHS.
6
Much of the material in this section is extracted from the pamphlet “Supercap 1 Farad Capacitor P6-8012”,
Copyright 1990 by Arbor Scientific.
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