Part 11. Capacitors
11.1
11.2
11.3
11.4
Capacitor Terminal Relationships
Power & Energy in Capacitors
Capacitors in Series & Parallel
DC Steady-State
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223
11.1 Capacitor Terminal Relationships
The capacitor is a physical device that is intentionally
designed to store energy in an electric field. Simply put,
capacitors store charge (q).
Capacitors are used for many purposes. Capacitors are used,
for example, as part of the circuits that tune in the stations
on your radio or television. Another example: In a portable
battery-operated photoflash unit, a capacitor accumulates
charge slowly during the charging process, building up an
electric field as it does so. It holds this field and its energy
until the energy is rapidly released during the flash.
Most capacitors consist of a pair of parallel aluminum plates
(conductors) electrically isolated from each other by an
insulator (a dielectric):
Conductor (aluminum)
Insulator (ceramic or other
dielectric material)
Conductor (aluminum)
Because of the insulating material between
the conductors, there is no flow of DC
current. Thus, a capacitor acts as an open
circuit in the presence of DC currents.
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assorted capacitors
ceramic disc caps
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225
i = dq/dt = conduction current (moving
charge)
+
+ + + +
v
_ - - - -
i = d/dt
= displacement current
(changing stored charge)
i = dq/dt = conduction current (moving
charge)
If the voltage v changes as a function of time, then the charge
that has accumulated on the plates will also change as a
function of time. For dielectrics, the ratio of flux to voltage
and the charge-to-voltage are constant; that is,
q = C v This is a kind of “Ohm’s Law”
The proportionality constant, C, measured in farads, is the
capacitance of the capacitor. It is a measure of how much
charge must be put on the plates to produce a certain voltage:
the greater the capacitance, the more charge is required.
Thus, even though no DC current is flowing through the
capacitor, a time-varying voltage causes charge to vary with
time. The change with time in the stored charge is analogous
to a current, (See p4 of Mayergoyz.)
i = dq/dt = Cdv/dt
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226
As a circuit element:
...
C +
v
–
i
Passive convention!
q = Cv
Memorize!
i(t) = C dv/dt
...
v(t) = 1/C i(t) dt
t
= 1/C i(t) dt + v(to)
to
Energy = 1/2 C v2 (J)
The voltage is proportional to the charge, which is the integral
of the current.
Note that this equation implies that v(t) cannot change
instantaneously. (So does the equation i = C dv/dt). The current can
change instantaneously, but the voltage cannot; it must be continuous
in time:
v
i
OK
t
t
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227
Example 1
Given v as
shown. Find i.
v
10 V
1 F
t
0
...
...
+ v –
1
2
3
i
Solution:
i = C dv/dt
= 1 x 10–6 dv/dt
= 10–6 x slope
Note that i changes
instantaneously!
i
10–5 A
–10–5 A
t
0
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1
2
3
228
Example 2
Given i(t) as shown. Find v(t)
assuming that v(0) = 0.
i(t)
100 F
1A
...
...
+ v
0
i
–
1
2
1
2
t
Solution:
v(t) = v(to) + 1/C  i(t) dt
104 V
v(t)
v(0) = 0 V
v(2) = 1/C x area
= 1/(100 x 10–6) x 1 0
= 104 V
t
Note that v does not change
instantaneously.
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229
1 k
Example 3
Given v(t) = 25 e–10t
Find: i(t), vR(t), vs(t).
+
v
–
i
vs +
–
80 F
Solution:
i(t) = C dv/dt = 80x10–6 d/dt (25e–10t)
= 80x10–6 x (–10) x 25 e–10t)
= – 0.02 e–10t A
vR(t) = 1000 i = 1000 x (–.02 e–10t)
= – 20 e–10t
vs(t) = vR (t) + v(t)
= – 20 e–10t + 25 e–10t
= 5 e–10t
Note that v, i, vR, and vs, all have the same form,
namely,
k e–10t.
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230
Example 4 Capacitive Voltage Divider
We have already studied resistor voltage divider
circuits. We can also make voltage divider circuits
with capacitors.
I C
Find: v2.
1
V
+
C2
Solution:
V2
–
The output voltage is
1 t
1 t
1 t
V2 
Idt and the input voltage is V 
Idt 
Idt
C 2 0
C1 0
C 2 0
If we divide these equations and simplify we get
V2 
C1
V
C1  C 2
The formula says that the output voltage V2 becomes larger
if C1 is made larger. This is the opposite of the resistive
voltage divider, where V2 becomes larger if the “other”
resistor is made larger.
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Proximity Switches
.
The electrical devices we use in our daily lives contain
many switches. Most switches are mechanical, and use an
actuator that is pushed, pulled, slid, or rotated, causing two
pieces of conducting metal to touch and create a short
circuit. Sometimes designers prefer to use switches without
moving parts, to increase the safety, reliability,
convenience, or novelty of their products. Such switches
are called proximity switches. Proximity switches can
employ a variety of sensor technologies. For example,
some elevator doors stay open whenever a light beam is
obstructed.
Another sensor technology used in proximity switches
detects people by responding to the disruption they cause in
electric fields. This type of proximity switch is used in
some table lamps that turn on and off when touched and
elevator buttons with no moving parts. The switch is based
on a capacitor, which is a device whose terminal
characteristics are determined by electric fields. When you
touch a capacitive proximity switch, you produce a
change in the value of a capacitor, causing a voltage
change, which activates the switch.
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232
Elevator Call Button
front view
side view
The elevator call button is a small cup into which the finger is
inserted. The cup is made of a metal ring electrode and a circularplate electrode that is covered with an insulating layer to insulate it
from the ring. When you insert your finger between the electrodes,
then, because your finger is much more conductive than the
insulating covering surrounding the electrode, the circuit responds
as though another electrode, connected to ground, has been added.
The result is a three-terminal circuit containing three capacitors.
C1
C1
capacitor model
C2
C3
model with finger touch
The actual values of the capacitors are in the range of 10 to 50 pF,
depending on the exact geometry of the switch, how the finger is
inserted, whether the person is wearing gloves, and so forth.
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233
Assume that the elevator call button is placed in the
capacitive equivalent of a voltage-divider circuit,as shown.
button
vs(t) +–
25 pF
+
v(t)
–
With no finger touch present, the circuit and the voltage v(t) become
button
C1
vs(t) +–
v(t) 
+
v(t)
–
C2
C1
vs (t)
C1  C2
With a finger touch present, the circuit and the voltage v(t) become
button
C1
vs(t) +–
C2
C2
C3
v(t) 
+
v(t)
–
C1
vs (t)
C1  C2  C3
When the button is not pushed vs(t) drops. The drop is detected by
the elevator’s control computer and ultimately results in the elevator
arriving at the appropriate floor. Lamp circuits are similar.
Part 11: Capacitors
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11.2 Power and Energy in Capacitors
Power:
p = vi
= C v dv/dt
Energy:
[ p in terms of v and i ]
[ p in terms of v ]
 w = 1/2 C v2 (J)
Memorize
Note that the energy cannot change instantaneously
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The Medical Defibrillator
The ability of a capacitor to store potential energy is the basis
of defibrillator devices, which are used by emergency
medical teams to stop the fibrillation of heart attack patients.
In the portable version, a battery charges a capacitor to a high
voltage, storing a large amount of energy in less than a
minute. The battery maintains only a modest voltage; an
electronic circuit repeatedly uses that voltage to greatly
increase the voltage of the capacitor. The power, or rate of
energy transfer, during this progress is also modest.
Conducting leads (“paddles”) are placed on the victim’s
chest. When a control switch is closed, the capacitor sends a
portion of its stored energy from paddle to paddle through the
patient. As an example, when a 70 F capacitor in a
defibrillator is charged to 5000 V, the energy stored in the
capacitor is
w = 1/2 C v2 = 1/2 (70 x 10–6 F)(5000 V)2 = 875 J.
About 200 J of this energy is sent through the patient during
a pulse of about 2.0 ms. The power of the pulse is
p = w/t = 200/(2 x 10–3) = 100 kW,
which is much greater than the power of the battery itself.
Part 11: Capacitors
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11.3 Capacitors in Series & Parallel
Capacitors in Series: For capacitors in series a voltage
applied across the series combination is the sum of the
resulting voltages across each capacitor and the same current
flows through each one.
+ v1– + v2 – + v3 –
+ vN –
i

v
+
–
C1
C2 C3
CN
Same current!
KVL:
v = 1/C1  idt + 1/C2  idt + … +1/CN  idt
= ( 1/C1 + 1/C2 + … + 1/CN )  idt
= 1/Ceq  i(t) dt
where
1/Ceq = 1/C1 + 1/C2 + … + 1/CN or
Ceq–1 = C1–1 + C2–1 + … + CN–1
Capacitors in series behave like resistors in parallel!
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237
Capacitors in Parallel: Capacitors are in parallel when a
voltage that is applied across their combination results in
that same voltage across each capacitor.
i2
i1
i
C1
C2
CN
iN +
v
–
Same
voltage
!
i = C1 dv/dt + C2 dv/dt + … + CN dv/dt
= ( C1 + C2 + … + CN ) dv/dt
Ceq = C1 + C2 + … + CN
Capacitors in parallel behave like R’s in series! (sort of..)
Capacitors in parallel add.
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238
Example 5 Find Ceq
24
40
4
32
Ceq
1
3
2
3
56
Solution:
24
Ceq
Ceq
40
4
32
1
3
3
56
24
72
40 // 32 = 72
3 // 4 // 2 = 9
( 9– 1 + 56 – 1 + 72 – 1) – 1
= 7
9
1
3
2
56
Part 11: Capacitors
239
Cont.
24
7
1 // 7 = 8
Ceq
1
3
24
Ceq
8
Ceq = (24–1 + 8–1 +3 –1 ) –1
= 2
3
Part 11: Capacitors
240
Example 6
Find C1eq, C2eq, v2, and v1.
v2 18
16
90 V
+
–
8
v1
36
6
C2eq
20
8
v
20
C1eq
Solution:
20 series 20 = 10, 10 // 8 = 18, 18 // 6 = 24
 C1eq = 24
18 series 36 = 12, 12 series 24 = 8, 8 // 8 = 16
 C2eq = 16
16
90 V
+
–
16
+
v2
v2 = 16/32 x 90 = 45 V
–
C2eq
Part 11: Capacitors
241
cont.
16 45 V 18
90 V
+
–
v1
36
6
8
C2eq = 16
20
8
C1eq = 24
12 v1
45 V
+
–
6
20
20
8
v
v
20
C1eq = 24
12
45 V +
–
v1
24
Capacitor Voltage Divider
v1 = [ 12 / (12+24) ] • 45 = 15 V
Part 11: Capacitors
242
11.4 DC Steady-State
A circuit is in the DC steady-state if all currents and
voltages are constant. (Recall that we are assuming
that all the independent sources are DC sources.)
When only resistors and independent and dependent
DC sources are present (no Cs or Ls) and there is no
switching, the circuit is always in the DC steadystate.
When the circuit does contain capacitors, the DC
steady-state corresponds to the circuit with all the Cs
replaced by open circuits:
+
+
i=0
v
i
v

–
–
The values of the voltages and currents are said to
be the values at t =  and are designated, for
example, vC() .
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243
Example 7
Find vC() and iC() , the steady-state values of
and vC and iC, respectively. Find energy stored
by capacitor. 6 
20V
+
vC
–
+
–
iC
1/20 F
4
Solution:
In the steady-state, the capacitor appears as an
open circuit. The “steady state circuit” is:
6
20V
+
–
+
iC()=0
vC()
–
4
vC() = 4/(4+6) x 20 = 8 V
iC () = 0 A
Energy = 1/2 C v2 = 0.5 (0.05) 82 = 6.4 J
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244
11.5 Capacitors Before and
After Switch Operation
The capacitor voltage immediately after a switch
is operated has the same value as it did
immediately before the switch is operated.
(Not necessarily so, however, for resistors and
for inductors).
Advice for RC circuits with switches:
Draw the circuit that applies immediately
before the switch is operated, and calculate
the capacitor voltage and any other variables
of interest.
Then draw the circuit that applies
immediately after the switch is operated (The
capacitor voltage will not have changed!), and
calculate the values of the variables at this
new time instant.
Part 11: Capacitors
245
Example 14 7.13 J2H
Find vC and iC @ 0– and 0+.
t = 0 The switch opens at t = 0.
6
20V
Solution:
+
vC
–
+
–
@ 0– (switch closed): 6 
20V
iC
1/20 F
4
The switch is closed at t = 0.
+
vC
–
+
–
4
iC
vC(0–) = 4 / (4+6) x 20 = 8 V
iC(0–) = 0 A (Same as example 12.)
@ 0+ (switch open):
20V
+
–
6
The switch is open at t = 0.
8V
+
vC(0+) +
–
–
iC(0+)
4
vC(0+) = 8 V (Same as vC(0–).)
iC(0+) = – 8/4 = –2 A (Discontinuous!)
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Example 15 7.15 J2H
Find i1 & i2 @ 0– and 0+.
4
t = 0
3
45V
+
–
i1
C
i2
8
24 
8
24 
Solution:
@ 0– (switch open):
4
3
45V
+
–
i1
i1 (0–)
+ vC –
i2
= i2 (0–) = 45/15 = 3 A
vC(0–) = 4 x 3 = 12 V
Part 11: Capacitors
247
cont.
original:
45V
+
–
4
t = 0
3
i1
C
i2
8
24 
@ 0+ (switch closed): 4 
3
45V
+
–
i1
i1(0+) =
i2 (0+)
+–
12 V
8
i2
(45 – 12) / (6+3)
8 // 24 = 6 
24 
= 33/9 A
= i1(0+) x (24/(8+24) = 11/4 A
Both i1(t) and i2(t) are discontinuous at 0, but
of course the capacitor voltage has no jumps.
Part 11: Capacitors
248
Example 16 J2H 7.35
Find dv1/dt and dv2/dt @ 0+
t=0
12 
t=0
+ +
8A
8
v1 v2
– –
1/2 F
6
6
1/4 F
Solution:
@ 0– :
8A
12 
i8
8
+
v1(0–)
–
+ 6
v2(0–)
–
6
i8 (0–) = [24 / (24+8) ] x 8 = 6 A (current division)
v1(0–) = 8 x i8 = 8 x 6 = 48 V
v2 (0–) = .5 x 48 = 24 V
Part 11: Capacitors
249
cont.
@ 0+:
8A
12 
i8
8 +
–
iC1
i12
48 V
+
–
24 V
i6
6
iC2
i8 (0+) = 48 / 8 = 6 A
i12 (0+) = (48 – 24) / 12 = 2 A
iC1 (0+) = – i8 – i12 = – 8 A
i6 (0+) = 24 / 6 = 4 A
iC2 (0+) = i12 – i6
= –2A
dv1/dt = 1/C1 x iC1
= [1 / (1/2) ] x (– 8 ) = – 16 V/s
dv2/dt = 1/C2 x iC2
= [1 / (1/4) ] x (– 2) = – 8 V/s
Part 11: Capacitors
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