Week 3

Capacitance & Inductance

Transitions

Try your hand at graphing these functions

Graph: y = ex

Graph: y = e-x

Graph: y = 1 - e-x

e x y2 1 22 2 42 3 82 4 162 5 322 6 642 7 1282 8 2562 9 5122 10 1024

1 2 3 4 5 6 7 8 9 10-100

100

300

500

700

900

1100

x

Y

y = ex

y = ex

e x y2 1 0.52 2 0.252 3 0.1252 4 0.06252 5 0.031252 6 0.0156252 7 0.0078132 8 0.0039062 9 0.0019532 10 0.000977

1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

X

Y

y = e-x

y = e-x

e x y2 1 0.52 2 0.752 3 0.8752 4 0.93752 5 0.968752 6 0.9843752 7 0.99218752 8 0.996093752 9 0.9980468752 10 0.999023438

y = 1 - e-x

1 2 3 4 5 6 7 8 9 100

0.10.20.30.40.50.60.70.80.9

1

X

Y

t

eyyyty )]()0([)()(

• y() is the final value.• y(0+) is the initial value.• Tau,, is the time constant.

First-order SystemHas the same equation

Time Constant

For RC

C in Farads

For LC

L in Henrys

Unit Symbol Power Name

Milli m −3 Thousandth

Micro μ -6 Millionth

Nano N -9 Billionth

Pico p -12 Trillionth

Femto f -15 Quadrillionth

Atto a -18 Quintillionth

• Capacitance is the ability of a body to store an electrical charge. 

• Capacitance C is given by • Gives the voltage/current relationship

RC Circuit – Initial ConditionsAn 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?

Time(s)

VC

Which path do you think it takes?

Voltage Across the Resistor - Initially

VResistor

t (sec)

eIf we assume the battery has NO internal resistance, the voltage across the resistor will be the EMF.

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

t (sec)

Imax=e/R

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

t (sec)

Vcape

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

t (sec)

Imax=e/R

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 BehaviorThe 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/AmpsUnit for Capacitance = Farad = Coulombs/Volts

!

1

SECONDS

SecondsCoulombs

CoulombsCxR

Sec

CoulombAmp

Amps

Coulombs

Volts

Coulombsx

Amps

VoltsCxR

The “Time” ConstantIt 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

)4(

)3(

)2(

)1()1(

)1()1(

)1()(

1

RCV

RCV

RCV

eRCV

eRCV

etV

RCRC

RCt

1RC 31RC2RC 4RC

e

0.63e

0.63e

0.86e

0.86e

0.95e

0.95e

0.98e

0.98e

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.

Steady StateTransient

State

ε is the full voltage of the source

Charging Functions

RCt

o

RCt

RCt

eItI

etV

eCtq

)(

)1()(

)1()(

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.

Charge and voltage build up to a maximum…

…while current fades to zero

Capacitor Discharge – Resistor’s VoltageSuppose now the switch moves

downwards towards the other terminal. This prevents the original EMF source to be a part of the circuit.

VResistor

t (sec)

e

At t =0, the resistor gets maximum voltage but as the capacitor cannot keep its charge, the voltage drop decreases.

Capacitor Discharge – Resistor’s Current

IResistor

t (sec)

I= /e R

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.

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

Icap

t (sec)

I= /e R

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.

The bottom line to take away….

RCt

o

RCt

RCt

eItI

etV

eCtq

)(

)1()(

)1()(

When charging a capacitorCharge and voltage build up to a maximum…

…while current fades to zero.

RC

t

RC

t

eItI

etV

o

o

)(

)( When discharging a capacitor

All three fade away during discharge. RC

t

eqtq o

)( RC

qI

oo

Time to charge 63% = time constant “tau” = τ = RC

Capacitor Circuit Operation

Capacitor Circuit OperationRecall the Circuit Representation

LINEAR Caps Follow the Capacitance Law; in DC

The Basic Circuit-Capacitance Equation

tCvtq c

Where

Q The CHARGE STORED in the Cap, Coulombs

C Capacitance, Farad

Vc DC-Voltage Across the CapacitorDiscern the Base Units for Capacitance

cCVQ

Volt

CoulombFarad

“Feel” for Capacitance• Pick a Cap, Say 12 µF Recall Capacitor Law

Now Assume That The Cap is Charged to hold 15 mC

Find V c

Solving for Vc

cCVQ

!!V! 1250

Coul/Volt12x10

Coul15x106

3

c

c

V

C

QV

Caps can RETAIN Charge for a Long Time after Removing the Charging Voltage

Caps can Be DANGEROUS!

Forms of the Capacitor Law• The time-Invariant

Cap Law If vC at − = 0, then the traditional

statement of the Integral Law Leads to DIFFERENTIAL Cap Law

dt

tdvC

dt

tdqti C

cCVQ

The Differential Suggests SEPARATING Variables

tCdvdtti C Leads to The

INTEGRAL form of the Capacitance Law

tv

v

t C

C

dzCdyyi

t

C dyyiC

tv1

If at t0, vC = vC(t0) (a KNOWN value), then the Integral Law becomes

t

tCC

t

t

t

C

dyyiC

tvtv

dyyiC

dyyiC

tv

0

0

0

1

11

0

Capacitor Integral Law • Express the VOLTAGE

Across the Cap Using the INTEGRAL Law

Thus a Major Mathematical Implication of the Integral law

If i(t) has NO Gaps in its i(t) curve then

Even if i(y) has VERTICAL Jumps:

The Voltage Across a Capacitor MUST be Continuous

An Alternative View

The Differential Application

t

CC dyyiCC

tqtv

1

tt

tC

tdyyi

Cttv

1limlim

00

tvttv CCt

0

lim

If vC is NOT Continuous then dvC/dt → , and So iC → . This is NOT PHYSICALLY possible

dt

tdvCti C

C

Capacitor Differential Law • Express the CURRENT

“Thru” the Cap Using the Differential Law

Thus a Major Mathematical Implication of the Differential Law

If vC = Constant Then

This is the DC Steady-State Behavior for a Capacitor

A Cap with CONSTANT Voltage Across it Behaves as an OPEN Circuit

dt

tdvC

dt

tdqti C

C

0Ci Cap Current

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 The Circuit Symbol From Calculus, Recall an Integral

Property

Compare Ohm’s Law and Capacitance Law

Capacitor Ohm

Now Recall the Long Form of the Integral Relation

Cv

CiNote The Passive Sign Convention

)()( tdt

dvCti c

C

t

CC dxxiC

tv )(1

)( RR

RR

Riv

vR

i

1

t

t

tt

dxxfdxxfdxxf0

0

0

0

)(1

)(1

)(t t

t

CCC dxxiC

dxxiC

tv

The DEFINITE Integral is just a number; call it vC(t0) so

t

t

CCC dxxiC

tvtv0

)(1

)()( 0

Capacitor Summary contConsider Finally the Differential Application

dt

tdvCti C

C

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 DefinedFrom 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 Inductors STORE electromagnetic

energy They May Supply Stored Energy

Back To The Circuit, But They CANNOT CREATE Energy

For a Linear Inductor The Flux Is Proportional To The Current Thru it

dt

dvL

dt

diLvLi L

LL

Inductance Sign Convention• Inductors Cannot Create

Energy; They are PASSIVE Devices

• All Passive Devices Must Obey the Passive Sign Convention

Inductor Circuit OperationRecall the Circuit Representation

Separating the Variables and Integrating Yields the INTEGRAL form

Previously Defined the Differential Form of the Induction Law

In a development Similar to that used with caps, Integrate − to t0 for an Alternative integral Law

dt

diLv L

L

t

LL dxxvL

ti )(1

)(

00 ;)(1

)()(0

ttdxxvL

titit

t

LLL

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)=1.2e-2t

vc(t) = 20-12e-2t

+

-

t = 0

10R

8v0.05 F

20v

i(t)

Summary

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

Circuit Behavior of Reactive Components

CONCLUSION: Inductance