Review

Part 3 of Course

Passive Circuit Elements

v iR

1i v

R

div L

dt

1i vdt

L

1v idt

C

dvi C

dt

+

-

Rv

+

-

Lv

+

-

Cv

ii

i+

-

Energy stored in the capacitor

The instantaneous power delivered to the capacitor is

( )t tdv

w p t dt C v dt C vdvdt

( )dv

p t vi Cvdt

The energy stored in the capacitor is thus

21( ) joules

2w Cv t

Energy stored in the capacitor

221 ( )( )

2 2

q tw Cv t

C

q CvAssuming the capacitor was uncharged at t = -, and knowing that

represents the energy stored in the electric field established between the two plates of the capacitor. This energy can be retrieved. And, in fact, the word capacitor is derived from this element’s ability (or capacity) to store energy.

Parallel Capacitors

i1i 2i Ni

1C 2C NCv+

-

ii

eqCv+

-

1 1

dvi C

dt 2 2

dvi C

dt N N

dvi C

dt

1 2 1 2N N eq

dv dvi i i i C C C C

dt dt

1

N

eq kk

C C

Thus, the equivalent capacitance of N capacitors in parallel is the sum of the individual capacitances. Capacitors in parallel act like resistors in series.

Series Capacitors

11

1v idt

C

1 21 2

1 1 1 1N

N eq

v v v v idt idtC C C C

1

1 1N

keq kC C

The equivalent capacitance of N series connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitors. Capacitors in series act like resistors in parallel.

DC

1v 2v Nv

1C 2C NC

vi

+ + + -- -

i

eqCv+

-

DC

22

1v idt

C

1N

N

v idtC

Energy stored in an inductor

The instantaneous power delivered to an inductor is

( ) ( )t t

L

diw t p t dt L i dt L idi

dt

( )di

p t vi Lidt

The energy stored in the magnetic field is thus

21( ) ( ) joules

2Lw t Li t

Series Inductors

1 1

div L

dt 2 2

div L

dt N N

div L

dt

1 2 1 2N N eq

di div v v v L L L L

dt dt

1

N

eq kk

L L

The equivalent inductance of series connected inductors is the sum of the individual inductances. Thus, inductances in series combine in the same way as resistors in series.

DC

1L 2L NL

1v 2v Nv+ + + -- -

v

i

i

eqLv+

-

DC

Parallel Inductors

11

1i vdt

L

1 21 2

1 1 1 1N

N eq

i i i i vdt vdtL L L L

1

1 1N

keq kL L

The equivalent inductance of parallel connected inductors is the reciprocal of the sum of the reciprocals of the individual inductances. Thus inductances in parallel combine like resistors in parallel.

22

1i vdt

L

1N

N

i vdtL

1i 2i Ni

1L 2L NLv+

-

i

i

v+

-

eqLi

Complex Numbers

real

imag

A

x jy A

jAe A

1tany

x

cos sinje j Euler's equation:

cos sinA jA A

2 2A x y

cosx A

siny A

ComplexPlane

measured positivecounter-clockwise

cos sinj j je e e j

Note:

A

Comparing Sinusoids

sin 45 cos 135t t

sin cos2

t t

cos cost t

sin sint t

Note: positive angles are counter-clockwise

cos sin2

t t

cos sin by 90t t leads

cos - sin by 90t t lags

sin t

Re

Im

cos t

-sin t

-cos t

sin t

Re

Im

cos t

45

45

135

cos 45 cos by 45 and sin by 135t t t leads leads

AC Circuits and Phasors

jM MX e X X

XM

Recall that when the assumed form of the current j te cancelled out.

We were then left with just the phasors

A phasor is a complex number that represents the amplitude and phase of a sinusoid.

( ) j tMi t I e

was substituted into the differential equations, the

Impedance

Impedancephasor voltage

phasor current

VZ

I

2 2 2Z R L

R j L

V

I

zR j L Z V

ZI

1tanz

L

R

Note that impedance is a complex number containing a real, or resistive component, and an imaginary, or reactive, component.

Units = ohms

AC

R

LcosMV t

i(t)

+

-

VR

VL

+ -

-

+

R jX Z

resistanceR

reactanceX

Admittance

Admittance1 phasor current

phasor voltage

IY = =

Z V

2 2 2

RG

R L

R j L

V

I

2 2 2

1 R j LG jB

R j L R L

I

Y =V

conductance

Units = siemens

susceptance2 2 2

LB

R L

AC

R

LcosMV t

i(t)

+

-

VR

VL

+ -

-

+

Re

Im

Re

Im

Re

Im

V

V

V

V

V

V

I

I

I

I

I

I

I in phasewith V

I lags V

I leads V

Z1

DC

Z2

Z3 Z4I1

I2V

I

Zin

3 41 2

3 4in

Z Z

Z Z ZZ Z

We see that if we replace Z by R the impedances add like resistances.

Impedances in series add like resistors in series

Impedances in parallel add like resistors in parallel

Voltage DivisionZ1

DCZ2

+

VI

V1

V2

+

-

-1 2

V

IZ Z

1 1V Z I

2 2V Z I

But

Therefore

11

1 2

Z

V VZ Z

22

1 2

Z

V VZ Z

Instantaneous Power

( ) cosM vv t V t

( ) cosM ii t I t

( ) ( ) ( ) cos cosM M v ip t v t i t V I t t

( ) cos cos 22

M Mv i v i

V Ip t t

Note twice the frequency

Average Power

0 0

0 0

1 1( ) cos cos

t T t T

M M v it tP p t dt V I t t dt

T T

0

0

1cos cos 2

2

t TM M

v i v it

V IP t dt

T

2T

1 cos2 M M v iP V I

0v i 90v i Purely resistive circuit Purely reactive circuit

12 M MP V I 1 cos 90 02 M MP V I

Effective or RMS ValuesWe define the effective or rms value of a periodic current (voltage) source to be the dc current (voltage) that delivers the same average power to a resistor.

0

0

2 21( )

t T

eff tP I R i t Rdt

T

0

0

21( )

t T

eff tI i t dt

T

eff rmsI I root-mean-square

0

0

2 21 ( )t Teff

t

V v tP dt

R T R

0

0

21( )

t T

eff tV v t dt

T

eff rmsV V

Effective or RMS Values

( ) cosM vv t V t

1

22

01 1 cos 2 22 22rms M vV V t dt

2 1 1cos cos 22 2 2T

0

0

21( )

t T

rms tV v t dt

T

11 2 2

22

00

1

2 2 2 2 2M

rms M M

VtV V dt V

Using and

Ideal Transformer - Voltage

1 1( )d

v t Ndt

2 2( )d

v t Ndt

1 1 1

2 2 2

dv N Ndt

dv N Ndt

22 1

1

Nv v

N

11

1( )v t dt

N

The input AC voltage, v1, produces a flux

This changing flux through coil 2 induces a voltage, v2 across coil 2

1v 2v

2i1i

+ +

- -2N1NAC Load

Ideal Transformer - Current

12 1

2

Ni i

N

The total mmf applied to core is

NiF

Magnetomotive force, mmf

1 1 2 2N i N i F R

For ideal transformer, the reluctance R is zero.

1 1 2 2N i N i

1v 2v

2i1i

+ +

- -2N1NAC Load

Ideal Transformer - Impedance

11 2

2

N

NV V

Input impedance

2

2L

VZ

I

21 2

1

N

NI I

1v 2v

2i1i

+ +

- -2N1NAC Load

Load impedance

1

1i

VZ

I

2

1

2i L

N

N

Z Z

2L

i n

ZZ 2

1

Nn

NTurns ratio

Ideal Transformer - Power

12 1

2

Ni i

N

Power delivered to primary

P vi

22 1

1

Nv v

N

1 1 1P v i

1v 2v

2i1i

+ +

- -2N1NAC Load

Power delivered to load

2 2 2P v i

2 2 2 1 1 1P v i v i P

Power delivered to an ideal transformer by the source is transferred to the load.

Force on current in a magnetic field( )q v F BForce on moving charge q -- Lorentz force

Current density, j, is the amount of charge passing per unit area per unit time. N = number of charges, q, per unit volume moving with mean velocity, v.

v t

Sj

j Nqv

( )N V q v F B

( ) V F j B

L

dQ Nq Sv ti j S

dt t

V S L

( ) L F i B

i BForce per unit length on a wire is

Rotating Machine

B

i

i

X

Force in

Force out

+-

brushes

commutator

B

i

iX

Force in

Force out

+-

X

r

B

2 cosw r

flux 2 cosA rl B Β

area 2 cosA lw l r

l

d

dt

emf flux 2 cos d rlt t

E s B

emf 2 sin 2 sinabe rl rlt

B B

a

b

Back emf

2 sinabe rl B

Back emf

abe

B

i

i

X

Force in

Force out

+-

brushes

commutator

Armature with four coil loops

XXX

X

abe

N

S

aE

aR

tV

aI

Motor Circuit

t a a aV E I R

a aE K

Power and Torque

d a a dP E I

/a t a aI V E R

d a aK I

/a t a aI V K R

2at ad

a a

KV K

R R

AC Nodal Analysis

For steady-state AC circuits we can use same the method of writing nodal equations by inspection that we learned for resistive circuits. To do so, we must replace resistances with impedances.

We solved Problem 4.31 in Class:

-j1

j2

2 0 A 4 0 A

V1 V2

Change impedances to admittances

-j1

j2

2 0 A 4 0 A

V1 V2

j1 SS

-j/2 S

Nodal Analysis for Circuits Containing Voltage Sources That Can’t be

Transformed to Current Sources

1. Assume temporarily that the current across each voltage source is known and write the nodal equations in the same way we did for circuits with only independent voltage sources.

2. Express the voltage across each independent voltage source in terms of the node voltages and replace known node voltages in the equations.

3. Rewrite the equations with all unknown node voltages and source currents on the l.h.s. of the equality and all known currents and voltages on the r.h.s of the equality.

We solved #4.33 in text in Class: Note: V2 = 10

assume I2

-j4

j26 45 A 10 0 V

V1 V2

AC

+

-I0 I2

-j/2 S

j/4 S

1/2 S

1

2

0.5 0.25 0 4.243 6.743

0.25 1 2.5

j j

j j

V

I

AC Mesh Analysis

For steady-state AC circuits we can use same the method of writing mesh equations by inspection that we learned for resistive circuits. To do so, we must replace conductances with admittances.

We solved Problem 4.38 in text in Class: Find I1 and I2:

6 0 A

12 0 V I1

I2AC AC

j1

-j1

+ +

- -

1

2

2 1 1 1 12

1 1 1 0 6

j j

j j

I

I

What happens if we have independent current sources that can’t be transformed in the circuit?

1. Assume temporarily that the voltage across each current source is known and write the mesh equations in the same way we did for circuits with only independent voltage sources.

2. Express the current of each independent current source in terms of the mesh currents and replace known mesh currents in the equations.

3. Rewrite the equations with all unknown mesh currents and voltages on the left hand side of the equality and all known voltages and currents on the r.h.s of the equality.

We solved Problem 4.40 in text in Class: Find I0:

2 0 A 6 0 V

I1I2

AC

-j1

+

-

j2

I0

V2

+

-

Assume you know V2

Note I2 = -2

1

2

3 1 0 2 4

2 2 1 4 4

j j

j j

I

V

Matlab Solution: Know how to program

1

2

1.414 81.87

7.376 12.53

I

V

AC Thevenin's Theorem

AC Thevenin's TheoremThevenin’s theorem states that the two circuits given below are equivalent as seen from the load ZL that is the same in both cases.

VTh = Thevenin’s voltage = Vab with ZL disconnected (= ) = the open-circuit voltage = VOC

LinearCircuit

b

a

inZ

I

ZL

b

a

ThVACZL

inZ

IZth

AC Thevenin's Theorem

ZTh = Thevenin’s impedance = the input impedance with all independent sources turned off (voltage sources replaced by short circuits and current sources replaced by open circuits). This is the impedance seen at the terminals ab when all independent sources are turned off.

LinearCircuit

b

a

inZ

I

ZL

b

a

ThVACZL

inZ

IZth

We solved Problem 4.40 using Thevenin's Thm. in Class:

0

8 2 8 2 8.246 14.04

1 1 2 2 3 1 3.162 18.43

j j

j j j

I

6 2(1 ) 8 2OC j j V

2 0 A 6 0 V

AC

-j1

+

-

j2

I0

2 0 A 6 0 V

AC

-j1

+

- VOC

+

-

-j1

1 1TH j Z

8 2j AC

-j1

+

-

j2

I0

0 2.608 32.47 I

AC Superposition

Superposition Principle

The superposition principle states that the voltage across (or the current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.

Steps in Applying the Superposition Principle

1. Turn off all independent sources except one. Find the output (voltage or current) due to the active source.

2. Repeat step 1 for each of the other independent sources.

3. Find the total output by adding algebraically all of the results found in steps 1 & 2 above.

Example Done in Class:

Note that the voltage source and the current source have two different frequencies. Thus, if we want to use phasors, the only way we've solved sinusoidal steady-state problems, we MUST use superposition to solve this problem. We considered each source acting alone, and then found v0(t) by superposition.

sin cos 90t t Remember that

+

-AC

0.2F 1H 2cos10t30sin 5t+

-

v0(t)

By superposition:

Example

+

-AC

0.2F 1H 2cos10t30sin 5t+

-

v0(t)

1 20 0 0( ) ( ) ( )v t v t v t

0 ( ) 4.631sin 5 81.12 1.05cos 10 86.24v t t t

v0(t) = Response due to voltage source + Response due to voltage source