Capacitor

d

A

V

qC

:capacitora of dimension physical theon dependsonly eCapacitanc

A circuit element that stores electric energy and electric charges

A capacitor always consists of two separated metals, one stores +q, and the other stores –q. A common

capacitor is made of two parallel metal plates.

Capacitance is defined as: C=q/V (F); Farad=Colomb/volt

Once the geometry of a capacitor is determined, the capacitance (C) is fixed (constant) and is independent of voltage V. If the voltage is increased, the charge will increase to keep q/V constant

Application: sensor (touch screen, key board), flasher, defibrillator, rectifier, random access memory RAM, etc.

Capacitor: cont.

• Because of insulating dielectric materials between the plates, i=0 in DC circuit, i.e. the braches with Cs can be replaced with open circuit.

• However, there are charges on the plates, and thus voltage across the capacitor according to q=Cv.

• i-v relationship:

i = dq/dt = C dv/dt

• Solving differential equation needs an initial condition

• Energy stored in a capacitor: WC =1/2 CvC(t)2

Capacitors in

V=V1=V2=V3

q=q1+q2+q3

321321 CCC

V

qqq

V

qCeq

parallel series

V=V1+V2+V3

q=q1=q2=q3

321

321

111

1

CCC

q

VVV

q

V

Ceq

Inductor

i-v relationship: vL(t)= LdiL/dt

L: inductance, henry (H)Energy stored in inductors

WL = ½ LiL2(t)

In DC circuit, can be replaced with short circuit

Sinusoidal waves

• Why sinusoids: fundamental waves, ex. A square can be constructed using sinusoids with different frequencies (Fourier transform).

• x(t)=Acos(t+)• f=1/T cycles/s, 1/s, or Hz =2f rad/s 2t / rad

=360 t / deg.

Average and RMS quantities in AC Circuit

01

0

T

dttxT

tx

It is convenient to use root-mean-square or rms quantities to indicate relative strength of ac signals rather than the magnitude of the ac signal.

rmsrmsavermsrms VIPV

VI

I ,2

,2

T

rms dttxT

x0

21

Complex number review

A

Ae

jA

ba

bj

ba

abajba

j

sincos

2222

22

Euler’s indentity

ab

11

2

1

2

1

2

1

11212121

22221111

21

21

21 ,

A

Ae

A

A

c

c

AAeAAcc

AeAcAeAc

j

j

jj

Phasor

How can an ac quantity be represented by a complex number?Acos(t+)=Re(Aej(t+))=Re(Aejtej )

Since Re and ejt always exist, for simplicity

Acos(t+) AejPhasor representation

Any sinusoidal signal may be mathematically represented in one of two ways: a time-domain form

v(t) = Acos(t+)

and a frequency-domain (or phasor) formV(j) = Aej

In text book, bold uppercase quantity indicate phasor voltage or currents

Note the specific frequency of the sinusoidal signal, since this is not explicit apparent in the phasor expression

AC i-V relationship for R, L, and C

Resistive Load Source vS(t) Asint

tR

A

R

vRi

tAtvv

R

SR

sin

sin

vR and iR are in phase

Phasor representation: vS(t) =Asint = Acos(t-90°)= A -90°=VS(j)

IS(jw) =(A / R)-90°

Impendence: complex number of resistance Z=VS(j)/ IS(j)=R

Generalized Ohm’s law VS(j) = Z IS(j)

Everything we learnt before applies for phasors with generalized ohm’s law

Capacitor Load

CjCj

jj

C

j

XjI

jVZ o

CC

CC

1

90

tCAdt

dqi

Cvq

tAv

CC

CC

C

cos

sin

90sin1

tC

AiC

ICE

VC(j)= A -90°

Notice the impedance of a capacitance decreases with increasing frequency

o

cC X

AjI 0

Inductive Load

tL

Adt

L

Ai

dt

diLv

tAv

L

LL

L

cossin

sin

90sin tL

AiL

Phasor: VL(j-90°IL(j)=(A/L) -180°

ZL=jL

ELI

Opposite to ZC, ZL increases with frequency

AC circuit analysis

• Effective impedance: example

• Procedure to solve a problem– Identify the sinusoidal and note the excitation frequency.

– Covert the source(s) to phasor form

– Represent each circuit element by its impedance

– Solve the resulting phasor circuit using previous learnt analysis tools

– Convert the (phasor form) answer to its time domain equivalent.

Ex. 4.21 P188

R1=100 R2=75 C= 1F, L=0.5 H, vS(t)=15cos(1500t) V.Determine i1(t) and i2(t).

Step 1: vS(t)=15cos(1500t), =1500 rad/s.Step 2: VS(j)=15 0Step 3: ZR1=R1, ZR2=R2, ZC=1/jC, ZL=jLStep 4: mesh equation

0)()( -

)()( )(

221

211

jIZZZjIZ

jVjIZjIZZ

RLCC

SCCR

2

21

2

2

1

21

)(0

)(

)(CRLCCR

SRLC

RLCC

CCR

RLC

CS

ZZZZZZ

jVZZZ

ZZZZ

ZZZ

ZZZ

ZjV

jI

0)()( -

)()( )(

221

211

jIZZZjIZ

jVjIZjIZZ

RLCC

SCCR

63.012.09.3512.08.4785500

0157.83687

8.4785500

63300575504450683005000456007500

4450683757.66100

01568375

)(7505.01500

7.667.66

101500

11

1

1

6

jI

jjj

jj

jjI

jjLjZ

jjjCj

Z

L

C

)(63.01500cos12.01 Atti

R1=100 , R2=75 , C= 1F, L=0.5 H, vS(t)=15cos(1500t) V