Experiments in Electricity
Charging and Discharging a Capacitor Experiment
804
CHAPTER 26
Capacitance and Dielectrics
I
n this chapter, we discuss capacitors — device
tors are commonly used in a variety of elec
used to tune the frequency of radio receiv
eliminate sparking in automobile ignition syste
electronic flash units.
A capacitor consists of two conductors sep
that the capacitance of a given capacitor depen
terial — called a dielectric — that separates the co
study;
value of an
ectric current
s with parallel
26.1
DEFINITION OF CAPACITAN
Consider two conductors carrying charges of
sign, as shown in Figure 26.1. Such a combinat
pacitor. The conductors are called plates. A po
the conductors due to the presence of the cha
difference is the volt, a potential difference is o
this term to describe the potential difference a
two points in space.
What determines how much charge is on t
voltage? In other words, what is the capacity o
particular value of !V ? Experiments show tha
pacitor 1 is linearly proportional to the potent
tors; that is, Q # !V. The proportionality const
2 We can
carrying charges of equal magnitude
butofof
sign,
as write this rel
ration
theopposite
conductors.
a combination of two conductors is
called
a
capacitor.
The
capacitance as follows:
13.5
sistors,
ng any circuit
mbination
of
I. INTRODUCTION
1.1. Capacitor
(a)
Consider two conductors
shown in Figure1. Such
conductors are called plates. A potential difference V exists between the conductors due to
Definition of capacitance
The capacitance
a capacitor is the ratio
the presence of the charges. Because the unit of potential difference
is the volt,C aofpotential
either conductor to the magnitude of the pote
difference is often called a voltage. We shall use this term to describe the potential
Q
difference across a circuit element or between two points in space.
C!
!V
er supply, the
ectric current
carrying wire,
–Q
th of the wire
1 F " 1 C/V
the wire is
urrent is the
per second
Note that by definition capacitance is always a po
tential difference !V is always expressed in Equ
cause the potential difference increases linear
Q /!V is constant for a given capacitor. There
capacitor’s ability to store charge and electric p
From Equation 26.1, we see that capacitanc
The SI unit of capacitance is the farad (F), wh
Faraday:
+Q
The farad is a very large unit of capacitance.
pacitances ranging from microfarads (10$6 F)
cal purposes, capacitors often are labeled “mF”
cromicrofarads or, equivalently, “pF” for picofa
1
the current
Figure 1. A capacitor consists of
y a voltage
he
electrical
Although the total charge on the capacitor is zero (beca
Figure 26.1 A capacitor consists
on one conductor as there is excess negative charge on the
of two conductors carrying charges
magnitude of the charge on either conductor as “the charg
of equal magnitude but opposite
2 The proportionality between !V and Q can be proved fro
two
sign.conductors carrying charges
points in space.
pacitor. The conductors are called plates. A potential difference !V exists between
What determines
how
plates
of a the
capacitor
for a given
the conductors
duemuch
to the charge
presence isofon
the the
charges.
Because
unit of potential
is the what
volt, a potential
difference
oftendevice
called afor
voltage.
We shall
use at a
age? In difference
other words,
is the capacity
ofisthe
storing
charge
this term
to describe
the potential
difference
across
a circuit of
element
or between
ticular value
of !V
? Experiments
show
that the
quantity
charge
Q on a catwo
points
in
space.
itor 1 is linearly
proportional to the potential difference between the conducWhat determines how much charge is on the plates of a capacitor for a given
s; that is, voltage?
Q # !V.
depends
on storing
the shape
and
In The
otherproportionality
words, what is the constant
capacity of the
device for
charge
at asepa2 that
show
quantity
charge
Q on
capacitor
linearly
to
particular
value
of
!V ?the
Experiments
show
that
thea quantity
a caon of theExperiments
conductors.
We
can
write ofthis
relationship
as Qofis"charge
C !VQproportional
ifonwe
define
1
the
potential
difference
between
the
conductors.
The
proportionality
constant
depends
on
is linearly proportional to the potential difference between the conducacitance pacitor
as follows:
tors;shape
that is,
Q#
!V. The of
proportionality
constant
the shapeas
and
the
and
separation
the conductors.
We candepends
write thison
relationship
Q sepa=C ∆V if
2
we
define
capacitance
as follows:
ration
of the
conductors.
We can write this relationship as Q " C !V if we define
capacitance
as follows:
he capacitance
C of
a capacitor is the ratio of the magnitude of the charge on
ther conductor to the magnitude of the potential difference between them:
The capacitance C of a capacitor is the ratio of the magnitude of the charge on
either conductor to the magnitude ofQ
the potential difference between them:
C!
!V
C!
Q
!V
(26.1)
(26.1)
te that byNote
definition
capacitance
is always a positive
quantity.
Furthermore,
the pothat by
by definition
capacitance isisalways
a positive
quantity.
Furthermore,
the the
po- poNote that
definition capacitance
always
a positive
quantity.
Furthermore,
tial difference
!V is always
in in
Equation
26.1
as apositive
positive
quantity.
tentialdifference
difference
isexpressed
alwaysexpressed
expressed
inEquation
Equation
26.1
quantity.
Be- Betential
V!V
is always
26.1
asasa apositive
quantity.
Because
cause
the potential
difference
increases
linearly
withthe
the stored
stored
charge,
the
ratio
se the potential
difference
increases
linearly
charge,
ratio
the
potential
difference
increases
linearly
with
the with
stored
charge,
the ratio
Q/V
is the
constant
Q
/!V
is
constant
for
a
given
capacitor.
Therefore,
capacitance
is
a
measure
of
a
for a given
Therefore,
capacitance
is a measure
of a capacitor’s
ability to store
!V is constant
forcapacitor.
a given
capacitor.
Therefore,
capacitance
is a measure
of a
capacitor’s
ability
to
store
charge
and
electric
potential
energy.
charge and electric potential energy.
acitor’s ability
toEquation
store charge
electric
potential
From
26.1, weand
see that
capacitance
has SIenergy.
units of coulombs per volt.
From Equation
26.1,
we we
seesee
that
hasSIwas
SI
units
of
coulombs
The SIEquation
unit
of capacitance
is that
thecapacitance
farad
(F), which
named
in honor
of Michael
From
26.1,
capacitance
has
units
of coulombs
per
volt.per
Thevolt.
SI
of capacitance isisthe
farad
(F), which
named
honor of Michael
Faraday:
Faraday:
e SI unit unit
of capacitance
the
farad
(F), was
which
wasin named
in honor
of Michael
1 F " 1 C/V
aday:
The farad is a very large unit of capacitance. In practice, typical devices have ca1 F " 1 C/V
pacitances ranging from microfarads (10$6 F) to picofarads (10$12 F). For practicapacitors
often
are labeled “mF”
for microfarads
and “mmF”
mi- caiscala purposes,
very large
unit of
capacitance.
In practice,
typical
devicesforhave
cromicrofarads
or, equivalently,
for picofarads.
Any
two conductors
separated by“pF”
an insulator
(or vacuum) form a capacitor.
A capacitor is
$6
$12
e farad
itances ranging from microfarads (10 F) to picofarads (10
F). For practia circuit element that accumulates charge when connected to a circuit. This accumulating
purposes,
capacitors
are
labeled
“mF”
for its
microfarads
and
“mmF”
for mi1 Although
charge
gives
riseoften
to a voltage
difference
V across
terminals
(plates).
mostcharge
practical
the total
charge
on the capacitor
is zero
(because
there
is as much
excess In
positive
onsists
applications,
each
conductor
initially
has zero
net
charge
and electrons
are
transferred
microfarads
equivalently,
“pF”
for charge
picofarads.
on oneor,
conductor
as there
is excess
negative
on the
other),
it is common
practice to
refer
to the
charges
magnitude
the charge on
conductor
as “the
charge charging
on the capacitor.”
from
one ofconductor
to either
the other.
This
is called
the capacitor. Then, the two
2 The proportionality between !V and Q can be proved from Coulomb’s law or by experiment.
conductors have charges with equal magnitude and opposite sign, and the net charge on
capacitor
whole
remains iszero.
hough thethe
total
chargeas
ona the
capacitor
zero (because there is as much excess positive charge
osite
ne conductor as there is excess negative charge on the other), it is common practice to refer to the
When we say that a capacitor has charge Q (or, a charge Q is stored on the capacitor), we
nitude of the charge on either conductor as “the charge on the capacitor.”
mean that the conductor at higher potential has charge +Qand the conductor at lower
e proportionality
!V and
potentialbetween
has charge
-Q. Q can be proved from Coulomb’s law or by experiment.
The electric field at any point in the region between the conductors is proportional to the
magnitude Qof charge on each conductor. It follows that the potential difference Vab
between the conductors is also proportional to Q .
In the simple act of charging or discharging a capacitor, we find a situation in which the
currents, voltages and powers do change with time.
ply Kirchhoff’s loop rule to loop bghab
contains the capacitor) to find the poacross the capacitor. We enter this poequation without reference to a sign
a Capacitor
e charge 1.2
on Charging
the capacitor
depends
of the potential difference. Moving
op, we obtain
$V cap " 3.00 V # 0
Resistor
Answer
11.0 V.
Exercise
Reverse the direction of the 3.00-V battery and answer parts (a) and (b) again.
Answer
(a)
(b) 30 &C.
I 1 # 1.38 A,
28.4 RC Circuits
I 2 # "0.364
A,
I 3 # 1.02 A;
$V cap # 11.0 V
Capacitor
28.4
RC CIRCUITS
+
R
R
–
–q
C
+q
I
Switch
So far we have been analyzing steady-state
circuits, in which
the current is conS
S
ε
ε
Battery
stant. In circuits containing capacitors, the current may vary in time. A circuit con(c)
(a)
(b) t < 0
taining a series combination of a resistor
and a capacitort >is0 called an RC circuit.
Figure 28.16
(a) A capacitor in series with a resistor, switch, and battery. (b) Circuit diagram
representing this system at time t % 0, before the switch is closed. (c) Circuit diagram at time
t $ 0, after the switch has been closed.
Charging a Capacitor
Figure 2. Charging a capacitor
difference that
across the
the resistor.
We have
the sign
conventions
discusseduncharged.
earlier
Let us assume
capacitor
inused
Figure
28.16
is initially
There is no
for the signs on ! and IR. For the capacitor, notice that we are traveling in the dicurrentrection
whilefrom
switch
S is plate
open
(Fig.
28.16b).
If represents
the switch
is closed
at t # 0, howthe positive
to the
negative
plate; this
a decrease
in
Let
us
assume
that
the
capacitor
in
Figure
2
is
initially
uncharged.
There
is
no
current
potential.
Thus, weto
useflow,
a negative
sign forup
this a
voltage
in Equation
28.11.
Note thatand the
ever, charge
begins
setting
current
in the
circuit,
capacitor
q andSI are
that is
depend
on time
opposed
to steady-state
while switch
is instantaneous
open
If thevalues
switch
closed
at t(as
=0,
however,
chargevalbegins to flow,
4
begins to charge.
Note thatcharged.
during charging, charges do not jump across the cathe capacitor
setting upues)
a as
current
in theis being
circuit, and the capacitor begins to charge.Note that during
We can
use Equation
28.11
find the initial
in the
circuit and the
pacitor charges
plates
because
theacross
gaptobetween
thecurrent
plates
represents
an open circuit. Incharging,
do notonjump
the
ca-instant
pacitor
the
maximum charge
the capacitor.
At the
the plates
switch isbecause
closed (t "
the between the
0),gap
stead,represents
charge
transferred
between
each
plate
and
itsthat
connecting
plates
open
circuit.
Instead,
is28.11
transferred
between
eachwire
platedue
and to the
charge onisan
the
capacitor
is zero,
and fromcharge
Equation
we find
the initial
current
inestablished
the
circuit
I 0 is electric
ain
maximum
andestablished
is equal
to inbattery,
electric
fieldwire
thefield
wires
by the
capacitor
its
connecting
due
to the
the wiresuntil
by thethe
battery,
until theis fully
capacitor
fully
charged.
the plates
becomethe
charged,
the potential
difference
across
! charged,
charged.is As
the
plates As
become
potential
difference
across
the capaciMaximum current
(current at t " 0)
(28.12)
I0 "
the
increases.
The value
of
maximum charge
charge depends
on on
the the
voltage
of the of the
R the
torcapacitor
increases.
The value
of the
maximum
depends
voltage
battery. Once
the
maximum
charge
is
reached,
the
current
in
the
circuit
is
zero
because
this time,
potential difference
fromisthereached,
battery terminals
appears entirely
battery.AtOnce
thethemaximum
charge
the current
in the circuit is zero
the potential
difference
across
the capacitor
matches
thattosupplied
by the
across
the resistor.
Later, when
the capacitor
is charged
its maximum
valuebattery.
Q,
becausecharges
the potential
theis zero,
capacitor
that supplied by the
cease to flow,difference
the current inacross
the circuit
and the matches
potential difference
the battery
terminals appears
entirelyKirchhoff’s
across the capacitor.
Substituting
battery.
To
analyze
thisfrom
circuit
quantitatively,
let us apply
loop rule
to the circuit after the
I " 0 into Equation 28.11 gives the charge on the capacitor at this time:
switchTo
is closed.
Traversing
the loop
clockwise gives
analyze
this circuit
quantitatively,
let us apply Kirchhoff’s loop rule to the
(maximum charge)
(28.13)
Q " C!
circuit after the switch is closed. Traversing the loop clockwise
gives Maximum charge on the capac
To determine analytical expressions for the time dependence of the charge
and current, we must solve Equation 28.11 —qa single equation containing two vari" circuit
IR #must
0 be the same. Thus,
(28.11)
ables, q and I. The current in all parts of"
the series
C as the current flowing out of and
the current in the resistance R must be the same
into the capacitor plates. This current is equal to the time rate of change of the
where q/C
potential
difference
across
and
chargeisonthe
the capacitor
plates.
Thus, we substitute
into Equation
28.11IR is the potential
I "the
dq /dtcapacitor
and rearrange the equation:
1.3 Discharging a Capacitor
dq
q
4 In previous discussions of capacitors,
we
a steady-state situation, in which no current was
" assumed
#
dt
R
RC
!
!
present
in consider
any branch
the circuit
containing
Now weofare
considering
the caseanbefore the
Now
let us
theofcircuit
shown
in Figurea3,capacitor.
which consists
a capacitor
carrying
To find
an expression
for q, we
first
combine
thecharges
terms onare
the moving
right-hand
side:
steady-state
condition
is
realized;
in
this
situation,
and
a
current
exists
initial charge Q , a resistor, and a switch. The initial charge Q is not the same as in
thethe wires
connectedcharge
to the capacitor.
dq
C !discussion,
q
q # C!
maximum
Q in the previous
"
#
" # unless the dis- charge occurs after the
dt
RC
RC
RC
capacitor is fully charged (as described earlier). When
the switch is open, a potential
difference Q /C exists across the capacitor and there is zero potential difference across the
must be in order to be an exponent of e in Equations 28.14 and 28.15.
ion
/RC is shown
dimensionless,
it
the tcircuit
in Figureas28.18,
which consists of a capaciThe
energy
output
of the battery as the capacitor is fully charged is
ons
28.14
and
28.15.
al charge Q , a resistor, and a switch. The initial charge Q is
2. Afteristhe capacitor is fully charged, the
–Q
capacitor
is
fully
charged
energy stored
in the capacitor
Q
#
C
maximum charge Q in the previous discussion, unless the disR
C
he
energy
stored
in
the
capacitor
1
1
2
+Q
the capacitor
(as described
earlier).
is 2Qis fully
is just
halfWhen
the the
energy output of the battery. It is left as a
# charged
2 Cleft ,aswhich
put of difference
the battery.
a
tential
QIt/Cis exists
across
the capacitor and there is
(Problem
that the
energybegins
supplied
by
resistor
because
I#=0.to
0.If the
Ifshow
the
switch
is remaining
closed at t half
= 0, of
thethe
capacitor
to discharge
g half across
ofproblem
the the
energy
supplied
by I60)
rence
resistor
because
switch is closed
S
or.
the
resistor.
the
battery
as internal
in the
resistor.
tor begins
to through
dischargeappears
through
the
resistor. Atenergy
some time
t
!
!
!
!
t<0
e, the current in the circuit is I and the charge on the capaci(a)
). The circuit in Figure 28.18 is the same as the circuit in FigDischarging
a Capacitor
r the absence
of the battery.
Thus, we eliminate the emf !
8.18, which consists of a capaci1 to obtain the appropriate loop equation for the circuit in
a switch. Now
The initial
charge
Q is the circuit shown in Figure 28.18,
–q
let us
consider
which consists of a capaci–Q
evious discussion, unless the disR
I
C
R
C
q
+q
tor carrying
an0 the
initial charge
resistor, and a switch. The initial charge Q is
+Q Q , a (28.16)
(as described
earlier).
When
$
$ IR
#
C same
across the
capacitor
and there
is maximum charge Q in the previous discussion, unless the disnot
the
as the
use
If
the
switch
is
closed
I
#
0.
into thisoccurs
expression,
it becomes
I # dq /dtcharge
S
after
the capacitorS is fully charged (as described
earlier). When the
ugh the resistor. At some time t
t>0
t<0
switch
a potential difference
Q /C exists across(b)the capacitor and there is
dqis open,
q
I and the
charge
(a)
$R on
#the capacidt circuit
C indifference
zero
potential
across the resistor because I # 0. If the switch is closed
is the same
as the
FigFigure 28.18 (a) A charged caThus, we
eliminate
the
emf
!
at t #dq0, the 1capacitor begins to discharge pacitor
through
thetoresistor.
connected
a resistor andAt some time t
#$
dtin
oop equationFigure
for the
circuit
3.
Discharging
a
capacitor
a
switch,
which
is
open
t % 0.
RC
duringq the discharge,
the current
in the circuit is I and theatcharge
on the capaci–q
C
+Q
(b) After the switch is closed, a cur-
R
I 28.18 is the same as the circuit in FigC
tor is
(Fig.
28.18b).
The
circuit
in Figure
ression, using
theqsome
fact
thattime
at t #
gives
q # Qt during
0, the
rent that decreases
in magnitude
+q discharge,
At
the current
in
the
circuit
is I and the charge on
(28.16)
with
time
is
set
up
in
the direction
Rentech
t
ureqcapacitor
28.16
except
for
the
absence
of
the
battery.
Thus,
we
eliminate
the emf
dq
1is q . To obtain the appropriate loop
equation
for the
circuit
in Figure
3:
shown,
and the charge
on the
ca#$
dt
from
Equation
28.11
to
obtain
the
appropriate
loop
equation
for
the
circuit
in
pacitor
decreases
exponentially
, it becomes
q
RC
S
Q
0
#
!
#
Figure
q 28.18:t
! Q " # $ RC
(b)
ln
dt
Figure 28.18
Volume-2
III.
PROCEDURE
dt EXPERIMENTAL
RC
hat the capacitor is discharging
the current direc(28.17) is opposite
discharging capacitor
citor wasIntegrating
being charged.this
(Compare
the current
directions
in that
expression,
using
the fact
that both the
charge on the capacitor and the
e18b.)
gives We
the see
instantaneous
current
nentially at a rate characterized by the time constantq" dq
# RC.
1
#
(28.18)
Q q
Current versus time for a
#$
discharging capacitor
! "
ln
q
Q
–q
C
q
dq
1
#$
dt
s the initial current. The negative sign indicates that the
curq
RC
Charge versus time for a
Q $t /RC
e
RC
the
with time.
t>0
(28.16)
$ caIR # 0
(a)$
A charged
C
versus
time for a Procedure
5.
pacitor connected to a resistor
and Charge
Experimental
$t
/RC
(28.17)
q(t ) # Qe
t
discharging capacitor
a switch, which is open at t % 0.
Rentech
Experiments in Electricity
II.APPARATUS
(b) dq
After
the switch
closed,
a curWhen
we substitute I #
into isthis
expression,
it becomes
/dt
expression
the instantaneous
givesrespect to time gives rent
Q at t # 0,with
that decreases in current
magnitude
5.1. EXPERIMENTAL PROCEDURE
with time is set up in the
directionq
:
dq
Andcapacitors
then recalling that VR
IR , we get:
Resistance,
cables,
multimeter,
basic
electrical
set,
Charging
a Capacitor
2.2.
shown,
and the $R
charge
on the
ca#
dt
pacitor decreases exponentially
Part-4:
Charging
and
Discharging
a Capacitor
dt
C Current versus time for a
dq
d
Q with
time.
(28.18)
(t) #
#
(Qe $t /RC ) # $
e $t /RC
discharging capacitor
$
–Q
#$
RC
t
RC
q (t )
C
V
I (t ) R
t
0
dt I (t )
V
R
Figure 28.18
(17)pacitor connec
a switch, which
(b) After the sw
rent that decre
with time is set
shown, and the
(18) Figure-35: Switch
pacitor
decreas
set.
with time.
I (t ) , we get:
gives
q #Solving
Q at Eq.(17)
t # 0, for
#
+q
q(t )
RC
Since I (t ) is just the rate of change of q (t ) :
2.
ative sign indicates that the curng is opposite the current direcmpare the current directions in
q(t ) # Qe $t /RC
harge on the capacitor and the
Figure-14:
an initially uncharged capacitor. Switch
d by the time constant
" # Charging
RC .
I (t )
dq(t )
dt
(28.17)
Now turn
the switch
(19) Charge versus
discharging ca
switch (S)provided
closed,
the charge
on the up
Then
substituting
this into Eq.(18),current
we find:
Differentiating
this
expression
with
respect
to time
gives
the instantaneous
Set open.
up When
thethe
circuit
on the
side.
1) initially
Figure-34: Experimental set-up of the RC circuit for the
capacitor increases over time while current decreases.
as a function of time:
2.1.
Set
volt
2.2.
Mea
and discharging a capacitor.
dq(t )You
V can
q(tuse
)
2) If you have one multimeter, prepare it for 2charging
situations.
your multimeter
for
acro
(20)
measuring current and voltage.
dt
R
RC
2.3.
Now
Current
versus
dq
d
Q $t /RC
Figure-(14) showsI(t)
a simple
a ) # $
(28.18)
# circuit
# for charging
(Qe $t /RC
e
term
discharging
ca
1. Construct
dtthat
RC the RC circuit shown in the
A circuit
such dtas thisof
has supply.
a
3) capacitor.
Please make
the connection
power
circ
Figure-(34) to
obtain for
theq (t )
Equation-(20)
is aexperimentally
differential equation
cap
resistor and a capacitor in series is called an
charging
and
discharging
curves
of
the
where
theAmmeters
initial current.
The negative
sign
indicates
that
the
curQ /RC
# I 0 isthat
2.4.
Brin
4) Do
not forget
are connected
in
series
so
that
the
current
flows
through
and it can be shown that the solution for the
R C circuit. Initially the switch S is open and
current
Iis. given
rentthem.
direction
now that
the capacitor
is discharging
is Real
opposite
direc-internal
The ideal
ammeter
has a resistance
of zero.
ammeters
have some
(Fig
equation
by:the current
no current passes
through
theconnected
circuit. When
the
circ
resistance.
Voltmeters
are
in
parallel
to
resistive
elements
in
the
circuit
so
that
tion when the capacitor was being charged. (Compare the current directions in
Resistance
Capacitor
2.5.
Rec
measure
the
potential
across
(on charge
each side
thecapacitor
element. and the
Sand
t 0We
switch
is closed
at
, adfference
current
to the
I starts
Figs.they
28.16c
28.18b.)
see that
both
onof)the
(21)
q(t ) VC (1 e t / RC )
find
flow
into the
circuit, and theatcharge
starts to
R(k time
)
C (" F
current
decay
exponentially
a rateqcharacterized
by the
constant
#) RC.
accumulate on the capacitor. Now, we can
determine the behavior of I (t ), q(t ) and the
33
3.
1000
Here, the voltage of the power supply V , resistor
1.1.
Note
that
the
capacitor
used
in
this
For the
switch to
remove t
5.2.
LABORATORY REPORT
Table-10: The data values during th
Part-4: Charging and Discharging a Capacitor
capacitor.
Measured
t (sec)
5) In this experiment, the current flowing through a resis- tor will be measured asI (the
A)
RCplease
For the So
circuit fill
with
piece1of
voltage across the resistor1.is varied.
thea Tabe
forwire
this circuit.
.....
0
connected across the capacitor:
6) At time t=0 when we first close
the
switchand
S report
in thebelow
circuit,
capacitor has no charge,
1.1.
Measure
the the
current
and so the current I will be determined by the resistor alone. The capacitor here acts as a
the circuit.
I flowing
short circuit. At any later time, the charge
willinstart
to increase while the current decrease.
V instant
Measure
theVC
voltage
across the
Then, q (t) will reach the constant1.2.
value
of q =
. At this
the capacitor will be fully
charged. The current, on the other hand,
will
be power
zero at
this instant. At this step try to fill
terminals
of the
supply.
the Table 1.
Resistance
Capacitor
R( k )
C( F )
.....
.....
3.
The
data
values
t (sec
obtaine
discharging of the capacito
the data Table-(11).
Table-9: The current in the short circuit.
Table-11: The data values during the
Measured
Measured
Calculated
7) Find the experimental time
constant 𝜏 of the circuit from the I vs t graphs.
Find it from
capacitor.
V
I
Table1 Charging a capacitor
t(second)
5Measured
V
current(A)
.....
I
V
R
t (sec)
Table2 Discharging a capacitor
0
.....
Measured
current(A)
t(second)
0
0
.
.
.
.
.
.
I ( A)
.....
both the charging and discharging graphs. Then, compare experimental time constant 𝜏
with its theoretical value obtained by
𝜏 =RC
45
Ref.
1) Serway, R, Beichner,R. Physics for Scientists ans engineers with modern physics, Fifth
edition. 2000.
2) Rentech.Experiments in electricity, student guide. 2013.
t (sec