Week 3
Capacitance & Inductance
Transitions
Try your hand at graphing these
functions
Graph: y = ex
Graph: y = e-x
Graph: y = 1 - e-x
y = ex
1100
900
e
2
2
2
2
2
2
2
2
2
2
x
1
2
3
4
5
6
7
8
9
10
y
2
4
8
16
32
64
128
256
512
1024
y = ex
700
500
Y
300
100
-100 1
2
3
4
5 x 6
7
8
9
10
e
2
2
2
2
2
2
2
2
2
2
y = e-x
0.6
x
1
2
3
4
5
6
7
8
9
10
0.5
y
0.5
0.25
0.125
0.0625
0.03125
0.015625
0.007813
0.003906
0.001953
0.000977
0.4
y = e-x
Y
0.3
0.2
0.1
0
1
2
3
4
5 X 6
7
8
9
10
y = 1 - e-x
e
2
2
2
2
2
2
2
2
2
2
x
1
2
3
4
5
6
7
8
9
10
y
0.5
0.75
0.875
0.9375
Y
0.96875
0.984375
0.9921875
0.99609375
0.998046875
0.999023438
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
1
2
3
4
5
6
X
7
8
9
10
First-order System
Has the same equation
y(t )  y()  [ y(0  )  y()]e

t

• y(∞) is the final value.
• y(0+) is the initial value.
• Tau,𝜏, is the time constant.
Time Constant
For RC 𝑡 = 𝑅𝑒𝑞 × 𝐶𝑒𝑞
C in Farads
For LC 𝑡 =
Unit
Symbol
Power
Milli
m
−3
Thousandth
𝐿𝑒𝑞
Micro
μ
-6
Millionth
𝑅𝑒𝑞
Nano
N
-9
Billionth
Pico
p
-12
Trillionth
Femto
f
-15
Quadrillionth
Atto
a
-18
Quintillionth
L in Henrys
Name
• Capacitance is the ability of a body to store an
electrical charge.
• Capacitance C is given by 𝐶 =
𝑞
𝑉
• Gives the voltage/current relationship
RC Circuit – Initial Conditions
An RC circuit is one where you have a capacitor and
resistor in the same circuit.
Suppose we have the following circuit:
Initially, the capacitor is UNCHARGED (q = 0) and the current through the resistor is
zero. A switch (in red) then closes the circuit by moving upwards.
The question is: What happens to the current and voltage across the resistor
and capacitor as the capacitor begins to charge as a function of time?
Which path do you
think it takes?
VC
Time(s)
Voltage Across the Resistor - Initially
e
VResistor
If we assume the battery
has NO internal resistance,
the voltage across the
resistor will be the EMF.
t (sec)
After a very long time, Vcap= e, as a result the potential difference between
these two points will be ZERO. Therefore, there will be NO voltage drop
across the resistor after the capacitor charges.
Note: This is while the capacitor is CHARGING.
Current Across the Resistor - Initially
Imax=e/R
t (sec)
Since the voltage drop across the resistor decreases as the capacitor
charges, the current across the resistor will reach ZERO after a very
long time.
Note: This is while the capacitor is
CHARGING.
Voltage Across the Capacitor - Initially
Vcap
e
t (sec)
As the capacitor charges it eventually reaches the same voltage as the
battery or the EMF in this case after a very long time. This increase
DOES NOT happen linearly.
Note: This is while the capacitor is CHARGING.
Current Across the Capacitor - Initially
Imax=e/R
t (sec)
Since the capacitor is in SERIES with the resistor the current will
decrease as the potential difference between it and the battery approaches
zero. It is the potential difference which drives the value for the current.
Note: This is while the capacitor is CHARGING.
Time Domain Behavior
The graphs we have just seen show us that this process depends
on the time. Let’s look then at the UNITS of both the
resistance and capacitance.
Unit for Resistance = W = Volts/Amps
Unit for Capacitance = Farad = Coulombs/Volts
Volts Coulombs Coulombs
R xC 
x

Amps
Volts
Amps
Coulomb
1 Amp 
Sec
Coulombs
R xC 
 SECONDS !
Coulombs
Seconds
The
“Time”
Constant
It is clear, that for a GIVEN value
of "C”, for any value of “R” it
effects the time rate at which the
capacitor charges or discharges.
Thus the PRODUCT of R and C
produce what is called the
CIRCUIT Capacitive TIME
CONSTANT.
We use the Greek letter, Tau, for
this time constant.
The question is: What exactly is
the time constant?
Another way
to express R
Another way to
express Farads as
Coulombs/Volt
The “Time” Constant
The time constant is the time that it takes for the capacitor to reach 63%
of the EMF value during charging.
Let’s test our function
V (t )  e (1  e
t
RC
V (1RC )  e (1  e
e
0.95e
0.86e
)
 RC
RC
V (1RC )  e (1  e 1 ) 
V (2 RC )  0.86e
V (3RC ) 
V (4 RC ) 
)
0.98e
0.63e
Transient
State
0.63e
Steady
State
0.95e
1RC
2RC
31RC
4RC
0.98e
ε is the full voltage of the source
Applying each time constant produces the charging curve we see. For
practical purposes the capacitor is considered fully charged after 4-5
time constants( steady state). Before that time, it is in a transient state.
Charging Functions
q (t )  Ce (1  e
V (t )  e (1  e
I (t )  I o e
t
t
t
RC
RC
RC
)
)
Charge
and voltage
build up to a
maximum…
…while current
fades to zero
Likewise, the voltage function can be divided by another constant, in this case,
“R”, to derive the current charging function.
Now we have 3 functions that allows us to calculate the Charge, Voltage, or
Current at any given time “t” while the capacitor is charging.
Capacitor Discharge – Resistor’s Voltage
Suppose now the switch moves
downwards towards the other terminal.
This prevents the original EMF source
to be a part of the circuit.
e
VResistor
At t =0, the resistor gets maximum voltage but as the
capacitor cannot keep its charge, the voltage drop decreases.
t (sec)
Capacitor Discharge – Resistor’s Current
Similar to its charging graph, the current
through the resistor must decrease as the
voltage drop decreases due to the loss of charge
on the capacitor.
Ie/R
IResistor
t (sec)
Capacitor Discharge – Capacitor's Voltage
The discharging graph for the capacitor is
the same as that of the resistor. There WILL
be a time delay due to the TIME
CONSTANT of the circuit.
In this case, the time
constant is reached
when the voltage of
the capacitor is 37%
of the EMF.
Capacitor Discharge – Capacitor’s Current
Similar to its charging graph, the current
through the capacitor must decrease as the
voltage drop decreases due to the loss of charge
on the capacitor.
Ie/R
Icap
t (sec)
The bottom line to take away….
Time to charge 63% = time constant “tau” =
When charging a capacitor
τ = RC
q (t )  Ce (1  e
V (t )  e (1  e
I (t )  I o e
When discharging a capacitor
All three fade
away during
discharge.
t
V (t )  e o e
I (t )  I o e
q(t )  qo e
t
RC


t
RC
RC
)
)
Charge
and voltage
build up to a
maximum…
…while current
fades to zero.
t
RC
t
RC

t
RC
qo
Io 
RC
Capacitor Circuit Operation
Capacitor Circuit Operation
Recall the Circuit
Representation

LINEAR Caps Follow the
Capacitance Law; in DC
Q  CVc
Where
Q  The CHARGE STORED
in the Cap, Coulombs
C  Capacitance, Farad
Vc  DC-Voltage Across the
Capacitor

The Basic Circuit-Capacitance
Equation
qt   Cvc t 
Discern the Base Units for
Capacitance
Farad 
Coulomb
Volt
“Feel” for Capacitance
• Pick a Cap, Say 12 µF

Recall Capacitor Law
Q  CVc

Solving for Vc
Q
15x10 3 Coul
Vc  
C 12x10 6 Coul/Volt
Vc  1250 V!!!
Now Assume That The Cap is
Charged to hold 15 mC
Find V c
Caps can RETAIN Charge for a Long
Time after Removing the Charging
Voltage
Caps can Be
DANGEROUS!
Forms of the Capacitor Law
Q  CVc


Leads to DIFFERENTIAL Cap
Law
dvC t 
dqt 
i t  
C
dt
dt

The Differential Suggests
SEPARATING Variables
it dt  CdvC t 

Leads to The
INTEGRAL form of the
Capacitance Law
t
vC t 

vC    
 i y dy  C 
• The time-Invariant
Cap Law

dz
If vC at − = 0, then the traditional
statement of the Integral Law
1 t
vC t    i  y dy
C 
If at t0, vC = vC(t0) (a KNOWN
value), then the Integral Law
becomes
1 t0
1 t
vC t    i  y dy   i  y dy
C 
C t0
1 t
vC t   vC t0    i  y dy
C t0
Capacitor Integral Law
• Express the VOLTAGE
Across the Cap Using the
INTEGRAL Law

Thus a Major Mathematical
Implication of the Integral law
qt  1 t
vC t  
  iC  y dy
C
C 

The Voltage Across a Capacitor
MUST be Continuous
An Alternative View


If i(t) has NO Gaps in
its i(t) curve then
The Differential Application
dvC t 
iC t   C
dt
1 t  t
lim vC t  t   lim  i  y dy
t 0
t 0 C  

Even if i(y) has VERTICAL
Jumps:
lim vC t  t   vC t 
t 0

If vC is NOT Continuous then
dvC/dt → , and So iC → . This
is NOT PHYSICALLY possible
Capacitor Differential Law
• Express the CURRENT
“Thru” the Cap Using the
Differential Law
dv t 
dqt 
iC t  
C C
dt
dt

Thus a Major Mathematical
Implication of the Differential Law

A Cap with CONSTANT Voltage
Across it Behaves as an OPEN
Circuit

Cap Current
If vC = Constant Then
iC  0


This is the DC
Steady-State Behavior for a
Capacitor
Charges do NOT flow THRU
a Cap
– Charge ENTER or EXITS
The Cap in Response to
Voltage CHANGES
Capacitor Current
• Charges do NOT flow THRU a Cap
• Charge ENTER or EXITS The Capacitor in
Response to the Voltage Across it
– That is, the Voltage-Change DISPLACES the
Charge Stored in The Cap
• This displaced Charge is, to the
REST OF THE CKT, Indistinguishable
from conduction (Resistor) Current
• Thus The Capacitor Current is Called the
“Displacement” Current
Capacitor Summary
From Calculus, Recall an Integral
Property
The Circuit Symbol
iC

vC
Note The Passive
Sign Convention
Compare Ohm’s Law and
Capacitance Law
Capacitor
t
t0
t


t0
 f x dx   f x dx   f x dx
Now Recall the Long Form of the
Integral Relation

iC (t )  C
t
dvc
(t )
dt
1
vC (t )   iC ( x)dx
C 
t
Ohm
1
vR
R
vR  Ri R
iR 
t
1 0
1
vC (t )   iC ( x)dx   iC ( x)dx
C 
C t0
The DEFINITE Integral is just a
number; call it vC(t0) so
t
1
vC (t )  vC (t0 )   iC ( x)dx
C t0
Capacitor Summary cont
Consider Finally the Differential Application
dvC t 
iC t   C
dt
Some Implications
• For small Displacement Current dvC/dt is
small; i.e, vC changes only a little
• Obtaining Large iC requires HUGE Voltage
Gradients if C is small
Conclusion: A Capacitor RESISTS
CHANGES in VOLTAGE ACROSS It
CONCLUSION: Capacitance
The Inductor
• Second of the Energy-Storage Devices
• Basic Physical Model:
Circuit Symbol
Physical Inductor
• Inductors are Typically Fabricated by Winding Around a
Magnetic (e.g., Iron) Core a LOW Resistance Wire
– Applying to the Terminals a TIME VARYING Current Results in a
“Back EMF” voltage at the connection terminals
Some Real
Inductors
Inductance Defined
From Physics, recall that a
time varying magnetic flux,
, Induces a voltage Thru
the Induction Law


Where the Constant of
Proportionality, L, is called the
INDUCTANCE
L is Measured in
Units of “Henrys”, H
1H = 1 V•s/Amp
d
vL 
dt
For a Linear Inductor The Flux Is
Proportional To The Current Thru it
  LiL
diL
 vL  L
dt


Inductors STORE electromagnetic
energy
They May Supply Stored Energy
Back To The Circuit, But They
CANNOT CREATE Energy
Inductance Sign Convention
• Inductors Cannot Create
Energy; They are
PASSIVE Devices
• All Passive Devices
Must Obey the Passive
Sign Convention
Inductor Circuit Operation
Recall the Circuit
Representation
Separating the Variables and
Integrating Yields the INTEGRAL
form
t
1
iL (t )   vL ( x)dx
L 
Previously Defined the
Differential Form of the Induction
Law
diL
vL  L
dt
In a development Similar to that
used with caps, Integrate − to t0
for an Alternative integral Law
t
1
iL (t )  iL (t0 )   vL ( x)dx; t  t0
L t0
Drill Problem 4-13, pp. 260-261.
The 0.05F in P4-13 is initially
charged to 8v. At t = 0, a 20v
source is connected.
Determine the expressions for:
I(t) and vc(t) for t > 0
i(t)
+
t=0
20v
-
10R
8v
0.05 F
I(t)=1.2e-2t
vc(t) = 20-12e-2t
Summary
Circuit Behavior of Reactive Components
Reactive
Element
Initial Conditions t = 0+
Final Condition t = ∞
Stored Quantity?
Source?
No
Yes
DC
AC
Capacitor
Short Circuit
Voltage Source
Open Circuit
Short Circuit
Inductor
Open Circuit
Current Source
Short Circuit
Open Circuit
CONCLUSION: Inductance