Introduction to Electrical Systems Course Code: EE 111 Course Code: - - PowerPoint PPT Presentation

Introduction to Electrical Systems Course Code: EE 111 Course Code: EE 111 Department: Electrical Engineering Department: Electrical Engineering Instructor Name: B G Fernandes Instructor Name: B.G. Fernandes E‐mail id: bgf @ee iitb ac in E‐mail id: bgf @ee.iitb.ac.in

EE 111: Introduction to Electrical Systems B.G.Fernandes Thu, Aug 06, 2009

Lecture 7 1/21

Sub‐Topic:

• Sinusoidal steady state analysis

EE 111: Introduction to Electrical Systems B.G.Fernandes Thu, Aug 06, 2009

Lecture 7 2/21

REVIEW Pulse response of R C circuit Pulse response of R‐C circuit:

Vdc + VC - + VC

T τ <<

No sharp edges in VC Almost similar waveform can be

• btained in R‐L circuit

Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

Lecture 7 3/21

Problem:

a) ‘S’ is kept open for long time. ‘S’ is then closed at an instant of time considered to be t = 0 msec. Determine Vc & Vo at t= 0+ if R1=1kΩ, R2=1kΩ R3=2kΩ & Is= 5mA

Eq circuit when S is open:

1μF

Is charges ‘C’

i i i = + ∵

As & v i i ∴ ↑ ↑ ∴ ↓

s c R

i i i = + ∵

As , &

c R c

v i i ∴ ↑ ↑ ∴ ↓

h 0 i 5 4 20 i V ∴ At Steady state when ic= 0 & iR= 5mA

⇒ EE 111: Introduction to Electrical Systems B.G.Fernandes

when 0 is 5 4 20

c c

v i V ∴ = × =

Thu, Aug 06, 2009

Lecture 7 4/21

( 0 )

20 & 20 2 5 10

c t

v V V

−

=

∴ =

( 0 )

20 2 5 10

• t

v V

−

=

= − × =

OR Steady state Vc = V = isR and τ = RC

Vo at t = 0+?

• +

(0 ) (0 )

= = 20V

c c

v v

+

(20 10) 2 20 13.33 3

• V

V − × ∴ = − =

EE 111: Introduction to Electrical Systems B.G.Fernandes Thu, Aug 06, 2009

+ (0 )

3

• Lecture 7

5/21

b) For t ≥0, find the expressions for Vo(t) and Vc(t) Vc attains a new value governed by circuit parameters.

Equivalent circuit at steady state

At steady state,

qu a e t c cu t at steady state

= 10+5 1=15V&

c

i v = ×

For t ( )

t f i f

v v v v e τ

−

≥ = + −

10+5 1 15V& 10 5 3 25 & 3 sec

• c

v v V m τ × = + × = =

( ) 25 , 20

c f in f f in t

v v v v e v V v

−

+ = = 25 5 ( 10) 2 5 15 3 3

c t c

• c

v e v V v e

τ τ −

∴ = − − × = − = −

EE 111: Introduction to Electrical Systems B.G.Fernandes

3 3

• c

Thu, Aug 06, 2009

Lecture 7 6/21

c) ‘S’ is again opened at t = 3msec, sketch Vo for 0 ≤ t≤ 10msec 0 ≤ t≤ 10msec At t = 3msec, Vo=14.4V. When S is opened equivalent circuit is circuit is

vo

Vo→ 10 with τ= 4msec

t (msec) EE 111: Introduction to Electrical Systems B.G.Fernandes Thu, Aug 06, 2009

Lecture 7 7/21

Sinusoidal steady state analysis

Sinusoidal signal : Sinusoidal signal :

⇒ Waveform has periodic sinusoidal variation

2π repeats after p

sin( )

m

a A t ω α = +

Instantaneous value

a =

Peak value

m

A =

2 f ω π =

m

α = phase angle (radians)

2 f ω π =

f = frequency in Hz

ω = frequency in rad/sec

EE 111: Introduction to Electrical Systems B.G.Fernandes

f = frequency in Hz

Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

Lecture 7 8/21

R.M.S. or Effective Value

⇒ Steady value of current effective in converting power ⇒ Steady value of current effective in converting power

That current which produces same heating effect as that of DC

⇒

2

m rms

I I I = =

sin , 2 i

m

i I t I ω = Phasor Representation: 2 sin I t ω = Phasor Representation:

• In complex plane any vector can be

represented by

j

A Ae θ =

⇒

represented by

(cos sin ) A A j θ θ = +

Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

Lecture 7 9/21

⇒

ω If this phasor is rotating at an angular frequeny ( d/ )

( )

( )

j t

A t A e

θ ω +

=

(rad/sec)

[ ]

cos( ) sin( ) A t j t ω θ ω θ = + + +

If th l h t ti t d

⇒ ⇒

If there are several phasors rotating at same speed They are stationary w rt each other

⇒ ⇒ ⇒

They are stationary w.r.t. each other need not to appear in the

ω

representation d b h h l ‘ ’

Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

⇒ Represented by their phase angle at same ‘t’

Lecture 7 10/21

Phasor Diagram:

2 2 2

A A θ = ∠−

1 1 1

A A θ = ∠

( )

( ) sin v t V t ω θ +

&

( ) 2

m

V v t V θ θ ⎛ ⎞ = ∠ = ∠ ⎜ ⎟ ⎝ ⎠

( )

( ) sin

m

v t V t ω θ = +

rms value

2 ⎜ ⎟ ⎝ ⎠

10 30 I = ∠ −

&

⇒

230 60 V = ∠

10 2 sin 6 i t π ω ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ 6 ⎝ ⎠

230 2 sin 3 v t π ω ⎛ ⎞ = + ⎜ ⎟ ⎝ ⎠

Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

3 ⎜ ⎟ ⎝ ⎠

Lecture 7 11/21

Steady state analysis:

L t ( )

2 i I ( 2 )

j t

t V t V

ω

Let ( )

2 sin Im( 2 )

j t

v t V t Ve ω ω = =

V V = ∠

• r

is applied to series R‐L‐C circuit

2 sin( ) i I t ω θ = +

( )

Im ( 2 )

j t

Ie

ω θ +

=

I I θ = ∠

Im ( 2 ) Ie =

• r

be the current flowing in it

( )

R L C

v v v v t + + =

flowing in it

( )

R L C

R

v Ri =

( )

1

R

V IR = …

Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

Lecture 7 12/21

L

di v L j LI dt ω = =

2

1 2

j

j e

π

π = = ∠

dt

2

( ) 2

L

V L I π ω θ ∴ = ∠ + 2

( ) (2) 2

L

X I π θ = ∠ + …

1 1 ( ) ( )

c

v i t dt i t C j c ω = =

∫

C

( ) 2 = I ( ) (3) I I jI V X π θ π θ ∠ − − ∴ = = = ∠ − …

CI (

) (3) 2

c

V X j c c c θ ω ω ω ∴ ∠ …

Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes EE 111: Introduction to Electrical Systems B.G.Fernandes

Lecture 7 13/21

From eqn (1), ‘V’ and ‘I’ are in phase ‘V’ and ‘I’ are in phase From eqn (2), ‘V’ across the inductor leads ‘I’ by 900 q ( ) y

• r ‘I’ flowing in a inductor lags ‘V’ by 900

π 2

L

V LI π ω = ∠

π 2

L L

V X I π = ∠ 2 X fL Ω 2

L

X fL π =

Ω Inductive reactance

L

X

⇒ EE 111: Introduction to Electrical Systems B.G.Fernandes Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

L

Lecture 7 14/21

From equation (3) ‘V’ across the capacitor lags current by 900 by 90

2

C

I V C π ω = ∠ − 2

C

X I π = ∠ − 2 C ω 2

capacitive reactance

C

X

⇒

1 , 2

C

X fC π = Ω f

∴ In a series R‐L‐C circuit,

[ ( )]

L C

V I R j X X = + −

[ ] I R jX = +

Z R jX = +

[ ] I R jX = +

Z impedance

⇒

Z R jX +

impedance

2 2

Z R X = +

1

X tan R θ

− ⎛

⎞ = ⎜ ⎟ ⎝ ⎠

EE 111: Introduction to Electrical Systems B.G.Fernandes Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

Lecture 7 15/21

R ⎜ ⎟ ⎝ ⎠

θ

+ve in inductive circuit

θ

I d l

θ

‐ve in capacitive circuit &

V V = ∠

∴If

, Z Z θ = ∠

V I Z θ = ∠−

⇒

θ

Impedance angle Z cosθ power factor

⇒

If θ is +ve power factor (lag) inductive If θ is ‐ve power factor (lead) capacitive

Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

Lecture 7 16/21

In R‐L and series R‐L‐C circuit with XL>XC, 0

2 π θ < <

In R‐C and series R‐L‐C circuit with XC>XL, 0

2 π θ − > >

Ph di Phasor diagram:

Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

Lecture 7 17/21

Parallel circuit: d d itt

1

Z Impedance, admittance

1 Y Z = 1 Y =

2 2 2 2

R X j = −

G jB = −

' '

• Y

R jX +

2 2 2 2

j R X R X + +

G jB =

G Conductance +ve for inductive ckt

V

B Susceptance +ve for inductive ckt. ‐ve for capacitive ckt.

V I VY Z = =

(8 6) 10 36 Z j + ∠ (8 6) 10 36 Z j = + = ∠

1 0.1 36 Y = = ∠−

(0.6 0.8) j = −

Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

Z

( ) j

Lecture 7 18/21

Given &

1 2

V V =

Problem:

they are in quadrature

2 3 &

1200 I I P W = =

2 3

T k V f V2 should lead V1

⇒

Take V1 as reference

1 1

200 V jV ∴ + =

1

141.4 V V ∴ =

1 2

141.4 0 & 141.4 2 V V π ∴ = ∠ = ∠

2 3 1 C

I I R X = → = ∵

Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

Lecture 7 19/21

1

1 j V I R R ⎡ ⎤ + = ⎢ ⎥ ⎦ ⎣

3 2 1

&

C C

V V jV I I R jX X = = = −

• r

1 1

R R ⎦ ⎣

1 C C

R jX X 141.4 2 200 45 45 (1) I R R ∴ ∠ = = ∠ …

From vector diagram

1 1

R R

2 L

IR IX =

2 L

R X =

⇒

]

2 45 V I R X IR ⎡ ∠

2

141.4 90 2 45 IR ∠ = ∠

]

2 2 2

2 45

L

V I R jX IR ⎡ ∴ = + = ∠ ⎣

2

2

100 45 I R ∴ = ∠

Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

Lecture 7 20/21

From eq(1),

1

200 45 I R = ∠

1

R

1 2

2 R R =

2 2 1

1200 V I R

( )

2 2

141 4 (100)

2 1 2 1

1200, I R R + =

( )

2 2

141.4 (100) 1200 2R R + =

2

16.66 R = Ω

1

33.33 R = Ω

Thu, Aug 06, 2009 EE 111: Introduction to Electrical Systems B.G.Fernandes

Lecture 7 21/21