SAMPLE OF THE STUDY MATERIAL PART OF CHAPTER 3

Transcription

1 SAMPLE OF THE STUDY MATERIAL PART OF CHAPTER Introduction SYMMETRICAL COMPONENTS & FAULTS CALCULATIONS Most of the faults occurring naturally are unbalanced faults. Analyzing unsymmetrical faults via KCL and KVL are very cumbersome Unsymmetrical faults create unbalanced voltage and current phasors. In order to tackle the complexity of network for unbalanced phasors, 3 ϕ phasors can be decomposed into symmetrical components (Fortesque s Theorem ) These symmetrical components are called positive sequence, negative sequence and zero sequence. Each component is analyzed separately. Analysis of each component is easy as it is balanced set of phasors. The output for each component can be subsequently added to obtain final values. Final values will be unbalanced. Individual component responses can be added only when N/W for each sequence component are perfectly decoupled i.e. the result of one is independent of the other. Fortunately in power system, positive sequence N/W, negative sequence N/W and zero sequence N/W are not mutually coupled. This makes symmetrical component decomposition for analyzing unbalanced fault an effective tool * Three sets of balanced phasors which are the symmetrical components of three unbalanced phasors. Shown in below figures. Page : 189

2 Positive Sequence Components Negative Sequence Components Zero Sequence Components Vc1 V a1 V a2 V co V b V b0 V a0 V b1 V c2 Fig Components Zero Sequence Components Considering the positive sequence components, the vector V lags behind V by 120 electrical and the vector V leads the vector V by 120 electrical since these three are also equal in magnitude V = V = V V = V 120 V = V 120 * For the positive sequence components, Stator and rotor field directions are same. Considering the negative sequence components, the vector V leads V by 120 electrical, and the vector V lags behind V by 120 electrical. * Since these three are also equal in magnitude V = V = V V = V 120 V = V 120 * For the negative sequence components, Stator and rotor field directions are reverse. For zero sequence components, we have V = V = V Page : 190

3 Since the three un balanced vectors V, V, and V can be resolved into three sets of balanced vectors, the vector V is equal to the sum of the positive sequence component V of phase a, the negative sequence component of phase a, and zero sequence component V of phase a.and zero sequence component V of phase a. i. e., V = V + V + V * Since all power systems considered to be linear, so superposition principle holds good. Similarly V = V + V + V V = V + V + V 3 unbalanced vectors (V, V, V )resolved into a total of 9 vectors, i.e, V, V, V and V, V, V and V, V, V. Out of these, only 3 are unknown. So, only three components are linearly independent. (i.e. from the knowledge of the three vectors V, V, V we can find out all the remaining vectors) OPERATORS: An operator 'a' is introduced, which when operates upon a phasor rotates it by +120 without changing the magnitude of the phasor upon which it operates. It is represented as, a = = e = J0.866 a = = e = 0.5 J0.886 a = = 1 0 = e = a + a = 0 a = a; a ; = a ; a = 1; 1 a = 3 30 ; a 1 = a 1 = a = 3 30 a a = J 3; a + a = 1 a a = J 3 * Assuming phases a as the reference, the relationship between the symmetrical components of phases 'b' and 'c' in terms of symmetrical component of phase 'a' can be written. (Phase sequence is abc) V = V + V + V V = a V + av + V or in matrix form V = av + a V + V Page : 191

4 V a V a0 V b = 1 a 2 a V a1 V c 1 a a 2 V a2 V abc = [T] V 012 Where, T = symmetrical transformation matrix V a0 = (1/3) [V a + V b + V c ] V a1 = (1/3) [V a +a V b +a 2 V c ] V a2 = (1/3) [V a +a 2 V b +a V c ] or in matrix form V a V a1 = (1/3) 1 a a 2 V a2 1 a 2 a V a V b V c V 012 = [T -1 ] V abc where, T -1 = inverse symmetrical transformation matrix Current instead of Voltage: I a = I a1 + I a2 + I a0 I b = a 2 I a1 +a I a2 + I a0 I c = a I a1 + a 2 I a2 + I a0 or in matrix form I a I a0 I b = (1/3) 1 a a I 1 I c 1 a a I 2 I abc = [T] I 012 Where, T symmetrical transformation matrix I a0 = (1/3) [ I a + I b + I c ] I a1 = (1/3) [ I a +a I b +a 2 I c ] I a2 = (1/3) [ I a +a 2 I b +a I c ] I a0 I a1 = I a2 or in matrix form (1/3) 1 a a 2 1 a 2 a I a I b I c Page : 192

5 I 012 = [T -1 ] I abc where, T -1 = inverse symmetrical transformation matrix Average 3 phase power in Terms of Symmetrical Components: P = 3 [ V a0 I a0 cos θ + V a1 I a1 cos θ + V a2 I a2 cos θ ] Impedance: V abc [T] V 012 V 012 = [Z abc ] I abc = [Z abc ] [T] I = [T -1 ] [Z abc ] [T] I Z 012 = [T -1 ] [Z abc ] [T] (since Z 012 = V 012 / I 012 ) Z s Z m Z m [Z abc ] = Z m Z s Z m Z m Z m Z s Z s & Z m are self & mutual impedances Z [Z 012 ] = 0 Z Z Sequence Impedance and Networks: The impedance of a circuit when positive sequence currents alone are flowing is called the impedance to positive sequence current. Similarly, when only negative sequence currents are present, the impedance is called the impedance to negative sequence current. When only zero sequence currents are present, the impedance is called the impedance to zero sequence current. Let Z a, Z b, Z c, & Z n are three phase impedances and neutral impedance Z a0 = 1/3 (Z a + Z b + Z c ) zero sequence Z a1 = 1/3 (Z a +az b +a 2 Z c ) positive sequence Z a2 = 1/3 (Z a +a 2 Z b +az c ) negative sequence 3.2 Sequence N/Ws of a synchronous Generator: Positive sequence Network: The positive sequence network of a synchronous generator is a balanced and symmetrical three phase network with positive sequence generated voltages (E a1, E b1, E c1 ) in the three phase, Page : 193

6 symmetrical three phase star impedance Z, in each phase and positive sequence currents (I a1, I b1, I c1 ) flowing in the three phase of shown in the below figure. I a1 Z E C1 E a1 ~ ~ ~ E b1 + I b1 Z1 Z 1 I C1 Fig. 3.2 The three phase system can be replaced by a single phase network as shown in the below: I a1 + Z 1 E for the Generated voltage E a1 - ~ V a1 V for the Terminal voltage Fig. 3.3 The equation of the positive sequence network is V a1 = E a1 I a1. Z Negative sequence network: In any general synchronous generator which is designed to generate, balanced voltages the negative sequence generated voltages are always zero. The negative network containing negative sequence impedance Z 2 in each phase and negative sequence currents flowing through these carrying negative sequence voltage drops as shown below figure. Page : 194

7 I a2 Z 2 N Z 2 Z 2 I c2 I b2 Fig. 3.4 The three phase network can be replaced by a single phase network as shown in the below figure. Z 2 I a2 V a2 Fig. 3.5 The equation for the negative sequence network is V a2 = - I a2.z Zero sequence Network: The equation for the zero sequence network is V a0 = - I a0 Z 0 Z 0 I a0 V a0 Fig. 3.6 Page : 195

8 The equivalent circuit for the zero sequence networks is different from the positive and negative sequence networks the impedance offered to zero sequence currents depends on grounding of the star point. If the star point is not grounded, then the zero sequence impedance is infinity. If the star point is grounded through some impedance Zn, then the equivalent zero sequence impedance is equal to the sum of the zero sequence phase impedance and is equal to 3Z n. Balanced three phase system consists of positive sequence components only; the negative and zero sequence components being zero. The presence of negative or zero sequence currents in a three phase system introduces asymmetry and is indication of an abnormal condition of the circuit in which these components are found. The vector sum of the positive and negative sequence currents of an unbalanced three phase system is zero. The resultant solely consists of three zero sequence currents i.e., Vector sum of all sequence currents in three phase unbalanced system = I + I + I In a three phase, 4 wire unbalanced system, the magnitude of zero sequence component is one third of the current in the neutral wire. In a three phase unbalanced system, the magnitude of negative sequence component cannot exceed that of the positive sequence component. If the negative sequence component are greater, the phase sequence of the resultant system would be reversed. The current of a single phase load drawn from a three phase system comprises equal positive, negative and zero sequence component Sequence impedances of transformer for various configuration of 3-phase transformer connection Positive sequence and negative sequence impedances are same and equal to leakage inductance and winding resistance. However transformer zero sequence impedance depends upon the configuration of transformer connection. Configuration 1: Y Y transformer bank with any one neutral grounded. Page : 196

9 H L H reference bus L Z 0 Fig. 3.7 Configuration 2: Y Y transformer bank with both neutrals grounded. H L H L reference bus Z 0 Fig. 3.8 Configuration 3: Y transformer bank with grounded Y neutral. If the neutral of the star side is grounded, zero sequence currents can flow in star path and balancing zero sequence currents can flow in delta. But no zero sequence current can flow in lines of delta side. Page : 197

10 H L reference bus H Z 0 Fig. 3.9 L Configuration 4: Y transformer bank with ungrounded star, This is special case of configuration 3 where the neutral is grounded through Z =. Therefore no zero sequence current can flow in the transformer windings. H L reference bus H Z 0 Fig L Page : 198

11 Zero sequence impedance of an alternator is zero. If alternator is star connected and grounded with impedance Z then its zero sequence impedance is 3Z. Negative sequence impedance of an alternator is same as positive sequence impedance. But there is no negative sequence induced voltage E in an alternator. Positive sequence impedance and negative sequence impedance of transmission line are same. Zero sequence impedance of a transmission line is approximately three times its positive sequence impedance because of greater coupling. Example: The line to ground voltage on the high voltage side of a step up Transformer are 100 kv, 33 kv, 38 kv on phases a, b, & c respectively. The voltage of phase a leads That of phase b by 100 & lags That of phase c by Determine analytically The symmetrical component of voltage V = 100 0, V = , V = Solution: V = 1 3 (V + λ V + λ V ) 1 [ , ] J = V 1 3 [V + λ V + λ V ] 1 [ ] 3 V = J V = 1 3 (V + V + V ) J FAULT CALCULATIONS: Faults can be classified as two types: 1. Series faults 2. Shunt faults Shunt faults are characterized by increase in current and decrease in voltage, frequency and power factor. Page : 199

12 Series faults are characterized by decrease in current and increase in voltage, frequency and power factor. The series faults are classified as 1. One open conductor fault 2. Two open conductors fault The shunt type of faults are classified as: 1. Single line to ground fault 2. Line to Line fault 3. Double line to Ground fault 4. Three phase fault The first three faults are the unsymmetrical faults. The three phase fault is symmetrical faults. Severity & occurrence of Faults: power transfer slg II IIg 3- power Faults Time Fig Fault Severity Occurrence 1) 3- (LLL,LLLG) Severe 5% 2) Phase to phase ground (LLG) Severe 10% 3) Phase to phase fault (LL) Less Severe 15% 4) Single line to ground Faults (LG) Very less 70% * 1 cause of short circuit insulation failure. Overvoltage caused by lightning or switching surge. Insulation contamination salt spray, pollution. Mechanical causes overheating, abrasion. Page : 200

13 3.3.1 Faults on T.L. Most common lines are exposed to elements of nature 60 70% lightning stroke over voltage causes insulation to flash over Line to ground short circuit or line to line s.c. High winds topple tower, tree falls on line. Winds an ice loading mechanical failure of insulation. For, salt spray, dirty insulation conduction path insulation failure. S. C. other elements. Cabels (10 15%), C.B. (10 12%) generator, motor, X-mer (10 15%) much less common over loading for extended periods deterioration of insulation Voltage of the Neutral: The potential of the neutral when it is grounded through some impedance or is isolated, will be at ground potential unbalanced conditions such as unsymmetrical faults. The potential of the neutral is given as V n =I n Z n where Z n is the neutral grounding impedance and I n is neutral current. Here negative sign is used as the current flow from the ground to the neutral of the system and potential of the neutral is lower than the ground. For a 3- phase system, I n = I a +I b +I c = (I a1 + Ia 2 + Ia 0 ) + (a Ia 1 +aia 2 Ia 0 ) + (a I a1 + a 2 Ia 2 + Ia 0 ) = I a1 (1+a+a 2 ) + Ia 2 (1+a+a 2 ) +3Ia 0 = 3Ia 0 V n = -3I a0 Z n Since the positive sequence and negative sequence components of currents through the neutral are absent, the drops due to these currents are also zero Single Line to Ground Fault: 1. Most frequently occurring fault. 2. Usually assumed the fault on phase a for analysis purpose, phase b and phase are healthy. The boundary conditions are a) V a = 0.(1) b) I b = 0.(2) c) I c = 0.(3) Page : 201

14 I a Z ~ - + E a + ~ - E c - - ~ E b + Z I b Z The sequence network equations are V a0 = - I a0 Z 0..(4) V a1 =E a -I a1 Z 1 (5) V a2 = I a2 Z 2....(6) Fig I c The solution of these six equations will give six unknowns V a0, V a1, V a2 and I a0, I a1, and I a2 Since I a1 = (1/3) [I a + a I b + a 2 I c ] I a2 = (1/3) [I a + a 2 I b + a I c ] I a0 = (1/3) [I a + I b + I c ] Substituting the values of I b and I c from equations (2), (3)in above three equations I a1 = I a2 = I a0 = I a /3.(7) Equation (1) can be written in terms of symmetrical components V a = 0 = V a1 + V a2 + V a0 Now substituting the values of V a0, V a1 and V a2 from the sequence network equation, E a - I a1 Z 1 I a2 Z 2 I a0 Z 0 = 0 (8) Since I a1 = I a2 = I a0 Equation (8) becomes Page : 202

15 E a I a1 Z 1 I a1 Z 2 - I a1 Z 0 =0 I a1 = I f = I a = I a1 + I a2 + I a0 ; I = I = I and +ve, ve and zero sequence networks are connected in series. I f = Single Line of Ground fault with Z f : I a1 = ( ) ~ Z 1 I a1 V a1 Z 2 I a2 V a2 Z 0 I a0 V a0 Conclusion: Fig The three sequence networks are connected in series. 2. If the neutral of the generator is not grounded, the Zero network is open circuited. Line to Line Fault The line to line fault takes place on phases b and c The boundary conditions are I a = 0. (1) I b + I c = 0. (2) V b = V c...(3) Page : 203

16 a I a E a Z E C C Z Z E b b I b I c Fig The sequence network equations are V a0 = -I a0 Z 0.. (4) V a1 = E a = I a1 Z 1 V a2 = - I a2 Z 2 (5) (6) The solution of these six equations will give six unknowns V a0, V a1, V a2, and I ao, I a1 and I a2 Since I a1 = (1/3) [I a + a I b + a 2 I c ] I a = (1/3) [I a + a 2 I b + a I c ] I a1 = (1/3) [I a + I b + I c ] Substituting the values of I a, I b and I c from equations in above three equations I a1 = (1/3) [0 +a I b - a 2 I b ] = (1/3) [a- a 2 ] I b I a2 = (1/3) [0 +a 2 I b - a I b ] = (1/3) [ a 2 - a] I b I a0 = (1/3) [0 + I b - I b ] I a0 = 0 Which means for a line to line fault the zero sequence component of current is absent and positive sequence component of current is equal in magnitude but opposite in phase to negative sequence component of current, i.e., Page : 204

17 Z 1 I a1 Z 2 I a2 E a V a1 V a2 Fig I a1 = -I a2 (7) Since V b = a 2 V a1 + av a2 + V a0.. (8) V C = av a1 + a 2 V a2 + V a0 Substituting the equations (8) and (9) equations (3) a 2 V a1 + av a2 + V a0 = av a1 + a 2 V a2 V a0..(9) V a1 = V a2 (10) i.e., positive sequence component of voltage equals the negative sequence component of voltage. This also means that the two sequence networks are connected in opposition. Now making use of the sequence network equation and the equation (10) V a1 = V a2. E a - I a1 Z 1 = -I a2 Z 2 = I a1 Z 2 I a1 = I f = I b = - I c = a 2 I a1 + ai a2 + I a0 (I a2 = - I a1, I a0 =0) = (a a)ia = J 3 I a1 = Conclusion: 1. The connection of sequence currents are connected in parallel. 2. The phase difference between I a1 and I a2 for line to line fault should be 180 (I a1 = I a2 ). Line to Line fault with Z f I a1 = ( ) Double Line to Ground Fault: Here sequence networks are connected in parallel. Page : 205

18 Double line to ground fault takes place on phases b and c The boundary conditions are I a = 0.(1) V b = 0.(2) V c = 0.(3) The sequence network equations are V a0 = -I a0 Z 0.(4) V a1 = E a I a1 Z 1.(5) V a2 = -I a2 Z 2...(6) I a Z + E a Z E c E b Z I b C b I c Fig The solution of these six equations will give six unknowns V a0, V a1, V a2 and I a0, I a1 and I a2. Since V a0 = (1/3) [V a + V b + V c ] V a1 = (1/3) [V a + av b + a 2 V c ] V a2 = (1/3) [V a + a 2 V b + a V c ] Using above three equations and substituting for V a, V b and V c from the equation (2) and (3) V a0 = (1/3) [V a ] = V a /3 V a1 = (1/3) [V a +a. 0 +a 2. 0 ] = V a /3 Page : 206

19 V a2 = (1/3) [V a +a a.0 ] = V a /3 V a0 = V a1 = V a2.(7) Using this relation of voltages and substituting in the sequence network equations V a0 = V a1 Similarly I a0 Z 0 = E a I a1 Z 1 I a0 = [ ]..(8) V a2 = V a1 I a2 Z 2 = E a I a1 Z 1 I a2 = [ ] Now from equation (1) I a = I a1 + I a2 +I a0 =0.(9) E a + Z1 V a1 Z 2 V a2 Z 0 + I a2 + I a0 I a1 Substituting values of I a2 and I a0 from the equation(9) and (8) Therefore I a1 = I a1 = [ ] [ ] = 0 I f =I b + I c = 3 I a0 I a0 = -I a1 Fig Page : 207

20 I f = -3 I a1 Conclusion: Zero sequence and negative sequence networks are parallel and this is in series to the positive sequence. Double line to ground fault with Z f I a1 = 3.4 Three Phase Fault The boundary conditions are I a + I b + I c = 0 V a = V b =V c I a is taken as reference I b = a 2 I a and I c = ai a a I a E a Z E C Z Z E b C b I b I c Fig I a1 =1/3 (I a + a I b + a 2. I c ) and substituting the values of I b and I c I a1 = 1/3 (I a + a. a 2 I a + a 2 a I a ) Ia 1 = Ia I a2 = 1/3 (I a + a 2 I b + a I c ) = 1/3 (I a + a 4 I b + a 2 I c ) Page : 208

21 = 1/3 (I a + a I a + a 2 I c ) = (1 + a + a2 ) = 0 I a0 = 1/3 (I a + I b +I c ) = 0 V a1 = 1/3 (V a + a V b + a 2 V c ) = 1/3 (V a + a V a + a 2 V a ) = V a (1 + a + a 2 ) = 0 V a1 = 1/3 (V a + a 2 V b + a.v c ) = 0 Since V a1 = 0 = E a I a1 Z 1, I a1 = E a / Z 1 The frequent fault in transformer line is single line to ground fault. The most severe fault is 3-phase fault. The most severe fault nearer to generator is single line to ground fault. From the analysis of the various faults, the following observations are made: 1. Positive sequence currents are present in all types of faults. 2. Negative sequence currents are present in all unsymmetrical faults. 3. Zero sequence currents are present when the neutral of the system is grounded and the fault also involves the ground, and magnitude of the neutral current is equal to 3 I a0. When currents are entering a delta connected winding, positive sequence components of line currents lead the corresponding positive sequence components of phase currents (winding current) in delta winding by 90. Positive sequence components of line currents on the star side of a star delta transformer, lead the corresponding positive sequence line currents on the delta side by 90. Positive sequence components of line and phase voltages on the star side of a star delta transformer lead the corresponding positive sequence line and equivalent phase voltages respectively on the delta side by 90. Proof: Balanced three phase system consists of positive sequence components only; the negative and zero sequence components being zero. Balanced system, Voltages must be in the form of V b = a 2 V a V c = av a Page : 209

22 From Inverse symmetrical transformation, V 012 = [T -1 ] V abc V V = (1/3) V a a 1 a a V V V V = (1/3) 1 a a a V 1 a a a V = (1/3) 1 a a a V a 1 a a a 0 = 1/33 V a 1 + a + a 2 = 0 0 a 3 = 1, a 4 = a 3. a = a V V = V 0 0 V Load Flow So for balanced 3 ϕ system, V a0 = V a2 = 0; V a1 =V a * Load cannot be same for all time in the system. The power flow idea is to find out the voltage at different bus bar, sub - station, node point & the flow of power on these lines, with given constraints and specifications. Types of Buses: 1. Load bus: In this type of bus, P and Q are known. The unknowns are V and δ. 2. Slack Bus/Reference Bus/Swing Bus:- This bus is a special type of bus. Here real and reactive powers are not specified only V and δ are known. 3. Generator/Voltage control/pv Bus:- In this type of bus, P and V are known. Q and δ are not known. Page : 210

23 3.5 Power System Network P G1, Q G1 1 generator ~ ~ bus burs L 2 P L3, Q L3 P G3, Q G3 3 L 1 L 3 L 6 L 4 L P G2, Q G2 ~ P L2, Q L2 P L4, Q L4 P L5, Q L5 L Load Fig G Generator [Those bus who has self generator are called generated bus bar.] [Who s don t have generator called load bus bar.] Page : 211

24 to bus m to bus s to bus V k = V k δ P LK Q LK P GK gen Q GK Load P K = P GK P LK Q K = Q GK Q LK Where P injected power (real) in to bus. Fig & Q injected (reactive) power into bus. For load bus P = Q = 0 So we can write P = P (i) Q = Q (ii) On the behalf of equation (i) and (ii) load bus are drawing the power. Example: P km, Q km to bus m P kj, Q kj to bus J P ki, Q ki to bus i Bus K V k = V k δ P K Q K Fig Page : 212

25 P = P P P = P + P + P Q = Q Q Q = Q + Q + Q Static analysis of Power Network. Mathematical model of the Network T.L. nominal π model. Bus power injection S = V i = P + JQ. P = P Q, Q = Q Q. But Admittance matrix Taking small example P 1, Q 1 P 3, Q 3 1 L 2 3 L 1 L 3 2 P 2, Q 2 Fig Page : 213

26 Equivalent π model: P 3, Q 3 P 1 Q 1 Y 130 Y 310 I 1 3 Y 120 Y 13 Y 12 Y 23 Y 210 Y Y 230 Shunt admittance P 2, Q 2 Fig Now, power equation converting in current equation. So we can write equation. I = Y V + Y (V V ) + Y V + Y (V V ) I = Y V + Y (V V ) + Y V + Y (V V ) I = Y V + Y (V V ) + Y V + Y (V V ) I Y + Y + Y + Y Y Y (Y I = + Y + Y + Y ) V Y Y I Y V V Y Y (Y + Y + Y + Y ) I Y Y Y V I = Y Y Y V I Y Y Y V Y = Y = Y self admittance. (Driving point admittance) Y = Y = X transfer admittance or mutual adce. Y Page : 214

27 I = Y V V = Z I Z = [Y ] 1. Y formatting is so easy & & Z matrix comp. 2. Y is symmetric matrix Y = Y = Y Y = Y = Y Y = Y = Y Y matrix is sparse matrix. (Sparse means most of the elements of Y matrix are zero) Because there are 90% elements are zero. 3. Dimension of Y is (N N) N = No. of bus Power Flow Equation I = Y V I = Y V S = P + JQ = V I P + JQ = V Y V = V e. V Y = Y e, k, n = 1, 2, N. P + JQ = V Y P = V V Q = V V, K = 1, 2, N. V e ( ) Y cos (δ δ θ ) Y sin(δ δ θ ) Characteristics of power flow equation Power flow equation are algebraic static system. nonlinear Iterative solution Relate P, Q in terms of V, δ & Y elements P, Q f (V, δ) There are two methods to solve these non linear equations: Page : 215

28 1. Newton Raphson method 2. Gauss Seidal method Newton Raphson (N.R.) method has got quadratic convergence and is fast as compared to Gauss Seidal and always converge. But N.R. method requires more time per iteration. Gauss Seidal has got linear convergence, convergence is affected by choice of slack bus and the presence of sense capacitor. Example: Assume a three phase system with a sustained supply voltage of 2,300 volts from line to neutral and with line impedance Z = j 10Ω, Z = j10ω and Z = (10 + j 10)Ω. Compute the magnitude of fault current for a single line to ground fault at the end of the line. Solution: Supply voltage = 2300 V line/neutral Z = j 10 Ω Z = j 10 Ω Z = (10 + j 10)Ω For a single line to ground fault at a in Figure below. a I a b I b c I c I = 0, I = 0, V = 0 Fig then I = I = I V I = Z + Z + Z where V is prefault voltage 2300 = j10 + j j10 = j30 Also I = I + I + I = 3I = = A 10 + j30 Page : 216