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 Thu, Aug Lecture 7 B.G.Fernandes 1/21 06, 2009

Sub ‐ Topic: • Sinusoidal steady state analysis EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 2/21 06, 2009

REVIEW Pulse response of R C circuit Pulse response of R ‐ C circuit: V dc + V C - + V C τ << T No sharp edges in V C Almost similar waveform can be obtained in R ‐ L circuit EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 3/21 06, 2009

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 V c & V o at t= 0 + if R 1 =1k Ω , R 2 =1k Ω R 3 =2k Ω & I s = 5mA 1 μ F Eq circuit when S is open: I s charges ‘C’ = + = + ∴ ∴ ↑ ↑ ↑ ↑ ∴ ∴ ↓ ↓ ∵ ∵ As As v v , i i & & i i i i i i i i c R c s c R ⇒ At Steady state when i c = 0 & i R = 5mA ∴ ∴ = × = v when h i i 0 is 5 4 0 i 5 4 20 20 V V c c EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 4/21 06, 2009

∴ = v 20 V & − = c t ( 0 ) = − × = 20 20 2 5 2 5 10 10 v V V − = o t ( 0 ) OR Steady state V c = V = i s R and τ = RC V o at t = 0 + ? − × v = v = 20V (20 10) 2 ∴ = − = c - c + V 20 13.33 V (0 ) (0 ) o o 3 3 + + (0 ) EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 5/21 06, 2009

b) For t ≥ 0, find the expressions for V o (t) and V c (t) V c attains a new value governed by circuit parameters. At steady state, Equivalent circuit at steady state qu a e t c cu t at steady state = ≥ i 0 For t 0 c − t × × = + + − e τ = 10+5 1=15V& 10+5 1 15V& v v v v v v ( ( v v v v ) ) e c f f in i f f o = = = + × = τ = v 25 , V v 20 v 10 5 3 25 & V 3 sec m f in c − t ∴ = − τ v 25 5 e c − − × t ( v 10) 2 5 = − = − τ c V v 15 e o o c c 3 3 3 3 EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 6/21 06, 2009

c) ‘S’ is again opened at t = 3msec, sketch V o for 0 ≤ t ≤ 10msec 0 ≤ t ≤ 10msec At t = 3msec, V o =14.4V. When S is opened equivalent circuit is circuit is v o V o → 10 with τ = 4msec t (msec) EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 7/21 06, 2009

Sinusoidal steady state analysis Sinusoidal signal : Sinusoidal signal : ⇒ Waveform has periodic sinusoidal variation 2 π repeats after p = ω + α a A sin( t ) m a = Instantaneous value A = Peak value m m α = phase angle (radians) ω ω = = π π 2 f 2 f ω = frequency in rad/sec f = frequency in Hz f = frequency in Hz EE 111: Introduction to Electrical Systems Thu, Aug EE 111: Introduction to Electrical Systems Lecture 7 B.G.Fernandes 8/21 06, 2009 B.G.Fernandes

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 I = ω = = i I sin t , m I I m rms 2 = ω 2 sin 2 I I i t Phasor Representation: Phasor Representation: • In complex plane any vector can be represented by represented by ⇒ Ae θ = = θ + θ j (cos sin ) A A A j EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 9/21 06, 2009

⇒ ω If this phasor is rotating at an angular frequeny ( (rad/sec) d/ ) = θ + ω j ( t ) A t ( ) A e [ ] = ω + θ + ω + θ A cos( t ) j sin( t ) ⇒ ⇒ If there are several phasors rotating at same speed If th l h t ti t d ⇒ ⇒ They are stationary w rt each other They are stationary w.r.t. each other ⇒ ω need not to appear in the representation ⇒ Represented by their phase angle at same ‘t’ d b h h l ‘ ’ EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 10/21 06, 2009

Phasor Diagram: = ∠ θ = ∠− θ A A & A A 1 1 1 2 2 2 ( ( ) ) = ω ω + + θ θ v t v t ( ) ( ) V V sin sin t t rms value m ⎛ ⎞ V = ∠ θ = ∠ θ m v t ( ) ⎜ ⎜ ⎟ ⎟ V ⎝ ⎝ ⎠ ⎠ 2 2 ⇒ I = ∠ − V = ∠ 0 0 & 10 30 230 60 π ⎛ ⎞ = ω − ⎜ ⎟ i 10 2 sin t ⎝ ⎝ ⎠ ⎠ 6 6 π ⎛ ⎞ = ω + ⎜ ⎜ ⎟ ⎟ v 230 2 sin t ⎝ ⎝ ⎠ ⎠ 3 3 EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 11/21 B.G.Fernandes 06, 2009

Steady state analysis: Ve ω ω = ω = j j t t Let ( ) L t ( ) v t t 2 2 V V sin i t t Im( I ( 2 2 V ) ) = ∠ or is applied to series R ‐ L ‐ C circuit V V 0 = ω + θ i 2 sin( I t ) ω + θ = = j ( t ) Im ( Im ( 2 2 Ie Ie ) ) or = ∠ θ be the current I I flowing in it flowing in it + + = v v v v t ( ) ( ) R R L L C C = ( ) = … v Ri V IR 1 R R EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 12/21 06, 2009

π π di j = = ∠ = = ω 2 j e 1 v L j LI L dt dt 2 2 π ∴ = ω ∠ θ + 2 V L I ( ) L 2 π = ∠ θ + … X I ( ) (2) L 2 1 1 ∫ = = v i t dt ( ) i t ( ) ω c C j c π ∠ θ − I ( ) − π I jI 2 ∴ ∴ = = = ∠ ∠ θ θ − … … V V = X X C I ( I ( ) ) (3) (3) ω ω ω c C j c c c 2 EE 111: Introduction to Electrical Systems EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes B.G.Fernandes 13/21 06, 2009

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 90 0 q ( ) y or ‘ I’ flowing in a inductor lags ‘V’ by 90 0 π π = ω ∠ V LI L 2 π π = ∠ V X I L L 2 = π fL Ω Ω 2 2 X X fL L Inductive reactance ⇒ X L L EE 111: Introduction to Electrical Systems EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes B.G.Fernandes 14/21 06, 2009

From equation (3) ‘V’ across the capacitor lags current by 90 0 by 90 π π I = ∠ − = ∠ − V X I ω ω C C C C 2 2 2 2 1 = Ω capacitive reactance X , ⇒ X π C C 2 fC f = + − ∴ In a series R ‐ L ‐ C circuit, V I R [ j X ( X )] L C = = + + I R I R [ [ jX jX ] ] Z = + + Z Z R R jX jX ⇒ impedance impedance − ⎛ ⎞ X = + θ = 1 2 2 ⎜ ⎜ ⎟ ⎟ tan Z R X ⎝ ⎝ ⎠ ⎠ R R EE 111: Introduction to Electrical Systems Thu, Aug EE 111: Introduction to Electrical Systems Lecture 7 B.G.Fernandes 15/21 06, 2009 B.G.Fernandes

+ve in inductive circuit θ θ ‐ ve in capacitive circuit θ θ I Impedance angle d l ⇒ V = ∠− θ ∴ If = ∠ = ∠ θ 0 I & Z Z , V V 0 Z Z cos θ ⇒ power factor If θ is +ve power factor (lag) inductive If θ is ‐ ve power factor (lead) capacitive EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 16/21 06, 2009

π < θ < In R ‐ L and series R ‐ L ‐ C circuit with X L >X C, 0 2 − π > θ > In R ‐ C and series R ‐ L ‐ C circuit with X C >X L, 0 2 Ph Phasor diagram: di EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 17/21 06, 2009

Parallel circuit: 1 1 = admittance d itt Z Impedance, d � Y ' ' Z R X 1 = − Y = = = − j j Y G G jB jB + + + 2 2 2 2 2 2 2 2 R X R X R jX G Conductance +ve for inductive ckt +ve for inductive ckt. B Susceptance ‐ ve for capacitive ckt. V V = = I VY Z = + + = ∠ ∠ 0 Z Z (8 (8 j j 6) 6) 10 36 10 36 1 = − Y = = ∠− 0 (0.6 ( j j 0.8) ) 0.1 36 Z EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 18/21 06, 2009

Problem: Given & = V V 2 1 they are in quadrature = = I I 3 & P 1200 W 2 2 3 V 2 should lead V 1 ⇒ Take V 1 as reference T k V f ∴ + = V jV 200 1 1 ∴ = V 141.4 V 1 ∴ = ∠ = ∠ π V 141.4 0 & V 141.4 2 1 2 = → = ∵ I I R X 2 3 1 C EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 19/21 06, 2009

⎡ ⎤ V V jV 1 j = = = + = or I & I ⎢ ⎥ V I − 3 2 R ⎦ 1 R R jX jX X X ⎦ ⎣ ⎣ R R R 1 1 C C C C 1 1 141.4 2 200 ∴ ∠ = = ∠ … 45 45 (1) I R R R R 1 1 From vector diagram = = IR IX ⇒ R X 2 L 2 L ] ] ⎡ ⎡ ∴ = + = ∠ ∠ 0 0 V V I R I R jX X 2 2 IR IR 45 45 ⎣ 2 2 L 2 ∠ = ∠ 0 0 141.4 90 2 IR 45 2 2 100 ∴ = ∠ 0 I 45 R 2 EE 111: Introduction to Electrical Systems Thu, Aug Lecture 7 B.G.Fernandes 20/21 06, 2009