HOW GRAPHS AND CAP CIRCUITS FIT
At arbitrary time t:
Consider a resistor, an uncharged capacitor, a switch and a power supply all
hooked in series. Note also that the voltage across “a” and “b” is equal to
both the battery voltage and the sum of voltages across the resistor and
capacitor. That is:
C
R
i.) With the decrease of voltage across
the resistor, the current goes down.
Vo = VC + VR
=
C
e.) With the increase of voltage across
the plates, there is a decrease of
voltage across the resistor (remember,
the sum of the two must always
equal Vo ).
q
+ iR
C
Vo
i( t)
Vo
Vo
a
!VR = VR
!VC = VC
!VR " VR
!VC " VC
R
b
After a long period of time, or at time t = ! :
i.) The capacitor charges to its max.
Vo
ii.) With the capacitor completely charged, all the voltage drop is across
the capacitor, there is no voltage drop across the resistor and the current
in the circuit is ZERO.
The switch is thrown at t = 0. What happen in the circuit?
a.) The capacitor initially has no charge on its plates, so it provides no
resistance to charge flow in the circuit.
1.)
C
At = 0:
b.) With no charge on the capacitor
plates, there is no voltage across the
capacitor.
c.) That means that initially, all the
voltage drop is across the resistor.
i.) Using Ohm’s Law, and assuming
the initial current in the circuit is io , we
can characterize the initial situation as:
C
After a really long time:
R
R
e.) At t = ! , as all of the voltage is
across the capacitor, we can write:
!VR = io R
!VC = 0
3.)
0
Vo = VC + VR
= Q max / C
! Q max = CVo
Vo
io
Vo
!VR = 0
!VC = VC
Vo
i=0
Vo
0
Vo = VC + VR
= io R
V
! io = o
R
So what do the charge-on-capacitor and current as a function of time
graphs looks like?
At an arbitrary time t:
d.) As time proceeds, charge accumulates on the capacitor plates increasing
the voltage across the plates as it does.
2.)
4.)
Plate-Charge Characteristics of a Charging Capacitor
Characteristics of a Discharging Capacitor
charge on
capacitor
R
q = Q max
q = .87Q max
(
q ( t ) =Q max 1 ! e
! t /RC
In theory,
capacitor fully
charged at t =
)
!
q = .63Q max
Now consider an open switch, a resistor and a fully
charged capacitor in series with one another (initially,
there is no complete circuit as the switch is open). With
the capacitor charged, we can assume there is Q max
worth of charge on one plate and a voltage drop across
the cap of Vo.
switch thrown at t=0
with cap charged
C
At t = 0, the switch is closed putting the capacitor and
resistor in parallel.
no charge on
capacitor initially
q 0 =0
In that case, the resistor will see a voltage of Vo across its leads motivating a
current io to flow.
! = RC
2! = 2RC
(after time equal to one time constant,
the cap will have charged to .63 of its
max Q)
time
(after time equal to two time constants,
the cap will have charged to .87 of its
max Q)
As the charge flow diminishes from the capacitor, the current in the circuit
diminishes.
5.)
Current Characteristics of a Charging Capacitor
current
Plate-Charge Characteristics of a Discharging Capacitor
C
current
R
R
q ( t=0 ) =Q max = CVo
i0 =Vo /R
all voltage is
dropped across
resistor (none
across cap)
producing
maximum current
flow in circuit
7.)
i( t) =
(
(
d Q max 1 ! e ! t /RC
dt
))
" 1 %
= $!
(Q ) !e! t /RC
# RC '& max
"V %
= $ o ' e ! t /RC
# R&
= io e ! t /RC
(
i = .37i0
(
! = RC
(after one time constant’s worth of time,
the current will have dropped to .37 of
its maximum as capacitor charges)
switch thrown at t=0
)
Vo
)
i = .13i0
(before t = 0, all voltage
is across the capacitor
as the cap is fully
charged; when you
throw the switch, the
cap begins to dump
charge through the
resistor)
most voltage drop across
capacitor leaving little
voltage drop across
resistor and little current
flow in circuit
2! = 2RC
switch thrown at t=0
with cap charged
q ( t ) =Q max e ! t /RC
q(!) = .37Q max
q(2!) = .13Q max
charge on cap goes to
zero with time
! = RC
time
(after a time equal to one time
constant, the cap has lost .63 of its
charge leaving .37 of max Q)
(after two time constant’s worth of time,
the current will have dropped to .13 of
its maximum as capacitor charges)
6.)
C
2! = 2RC
time
(after a time equal to two time
constants, the cap has lost .87 of its
charge leaving .13 of max Q)
8.)
Current Characteristics of a Discharging Capacitor
current
R
as charge on cap decreases with it’s corresponding
voltage drop across the cap, current flow decreases with
the voltage across the resistor dropping appropriately
i0 =Vo /R
as far as the resistor is
concerned, the
charged cap initially
looks like a battery of
voltage
i( t) =
(
(
d Q max 1 ! e
! t /RC
dt
))
" 1 %
= $!
(Q ) !e! t /RC
# RC '& max
"V %
= $ o ' e ! t /RC
# R&
= io e ! t /RC
Q
Vo = max
C
(
i = .37i0
(
switch thrown at t=0
with cap charged
)
C
)
i = .13i0
as charge on cap goes to
zero, current goes to zero
! = RC
(after time equal to one time constant,
the current will have dropped to .37 of
its maximum as capacitor continues to
discharge)
2! = 2RC
time
(after time equal to two time constants,
the current will have dropped to .13 of
its maximum as capacitor continues to
discharge)
9.)