GATE CLASS 2025 - 2026 - POWER SYSTEM - Dr.L.UMASANKAR - ASSO.PROFESSOR - S.A. ENGINEERING COLLEGE.

追加: 2026-05-07

[00:00:01]Welcome to today's GATE video. I am Dr. El Umar Shankar, associate professor AC engineering college, electrical engineering department. So today going to explain about power system analysis subject. In the power system analysis the important topics symmetrical and unsymmetrical fault analysis.

[00:00:32]So what is meant by fault analysis. So fault analysis is nothing but study of abnormal operation in power system when an unended electrical connection occurs between the phases or between the phase and ground. So this is nothing but called as fault analysis.

[00:01:00]So what is the causes of faults? So different type of reasons are available. For example, cable breakdown is also create a fault.

[00:01:16]Flashover takes place in the conductor or insulator due to lightning initiation and tree falling and due to the construction activity and high-speed wind also create line-to-line fault and vehicle collision that is also damage the conductors that will create a fault.

[00:01:39]So these are the reason creating the faults.

[00:01:47]So what is the necessity of fault analysis?

[00:01:54]The purpose of fault analysis is designed the circuit breaker or otherwise selection of circuit breaker, relay protection design, equipment design, system stability studies and grounding system design and safety of personal equipment. Because during the fault huge amount of current will flow through the conductor.

[00:02:28]We know very well the power system network consist of many number of equipments that is costliest equipment. So to protect the costliest equipment we have to use some protective devices that is circuit breakers and switching gears. The circuit breakers depends the relay. Okay. So that we have to calculate the braking capacity of the circuit breaker and rating of the sushi gear also vary depends upon the fault current.

[00:03:06]Similarly relay protection design it's very important what is the pickup value of relay that is different with that is vary depends upon the nature of type of current clear and equipment design thermal rating of the conductor busar rating of the conductor mechanical strength of the conductor is also vary depends upon the fault type. Okay. Uh system stability studies we have to analyze behavior of the circuit. So what is the behavior of the circuit when any symmetrical fall current occurs or unsymmetrical fall current occurs? We will discuss later what is the difference between symmetrical fall current and unymmetrical fall current.

[00:04:00]So different type of fault currents are created or occurring in the power system network. So that analyzing the system behavior is also it's very important and grounding system design. What is the earthfall current value? Calculate the earthfall current and calculate the neutral grounding resistor value. That is also important and safety of personal and equipment. So due to the fall high amount of current will flow right that will create a flash or equipment explosion. So that so that only these are the parameters are important during the fault analysis.

[00:04:47]Okay.

[00:04:53]So different types of faults are available. So one is open circute fault.

[00:04:59]The another one is short circuit fault.

[00:05:02]So open circute fault different types that is one conductor is open or two conductor is open or three conductors are open. So what is going to be open circuit fault? So open circuit means open circute fault is nothing but a series fault.

[00:05:22]one or more conductors become open that is called as open circuit fault. So in this example the phase A is open. It may be due to the fuse blown out. It it may be fuse blown out or otherwise the conductor is damaged will create open circuit fault. So the effect of open circute fault is voltage is imbalanced and overheating is occurs in the rotating machine and reduce the system performance. So this is the effect of open circuit fault but the current value is decrease in open circuit fall compared with the short circuit fall. In short circuit the current value is very high. So uh all the prefaces are short circuited that nothing but triple line or otherwise triple line to ground is also class one of the classification of symmetrical fault. Okay. So in symmetrical fault all the fault current values are equal in all the phases in all the three phases so that it is a balanced one.

[00:06:41]Okay. And also the phase displacement between the phases also equal. So that only we can say it's nothing but symmetrical fault. So in this case in symmetrical fall we are using positive sequence network. So the occurrence of symmetrical fault is very rare but it is a most severe fault.

[00:07:09]The next one is unymmetrical fault.

[00:07:14]unsymmetrical fault classified as single line to ground fault, line-to-line fault and double line to ground fault. So single line to ground fault.

[00:07:37]So in so in this example Y Bphase short circuited with the ground is called as single line to ground fault.

[00:07:54]Lineto- line fault means two lines are short circuited. Two lines are short circuited is called as lineto- line fault.

[00:08:09]Next double line to ground fault means two lines are simultaneously grounded.

[00:08:19]Two lines are simultaneously grounded is called as double line to ground fault.

[00:08:29]In unsymmetrical case all the fall current values are not equal and also the phase displacement between the phases also not equal.

[00:08:43]So, so that only it is called as unbalanced fault. Unymmetrical fault is otherwise called as unbalanced fault. Okay. And one more point is important in symmetrical fault case all the three phases carrying the equal amount of fault current. Right? So that we can calculate if using any one of the if is equal to V tinian divided by E tinian plus EF.

[00:09:20]So what is the difference between open circuit fault is otherwise called as series fault. Short circuit fault is otherwise called shunt fault. So in open circuit current value is very less compared with the short circuit but in the short circuit current value is huge and it caused by broken the conductor here it caused by phasetophase or phase to ground contact and less severity open circuit fault is less severity that's what I mentioned less importance but short circuit fault is more severe and dangerous so there is a difference between open circuit fault and short circuit fault and frequency of occurrence. So in three-phase line to ground fault is nothing but a symmetrical type. So the probability of occurrence is 2 to 3 percentage. But in three-phase lineto- line fault uh again it is nothing but symmetrical case. Okay. Here the probability of occurrence is less than 1 percentage.

[00:10:32]In single line to ground fault is unsymmetrical case. This is frequently occurring but probability of occurrence is 70%age.

[00:10:44]Okay. Compared to a three-phase line to ground fault, it's nothing but uh severity is less but more more frequent frequently occur. Here line-to-line fault. So frequent probability of occurrence is 15 to 20%age. And double line to ground fault is also unymmetrical type. The frequency of occurrence is less than 10 percentage.

[00:11:11]So symmetrical fault is very rare but most severe but unymmetrical case single line to ground fault is less severe but frequency of occurrence is more. Okay that is the difference between unymmetrical and symmetrical fault.

[00:11:34]So this is a uh one of the example problem in symmetrical fault analysis.

[00:11:40]Symmetrical fault occurs on bus four of system determine fault current post fault wtages and line current. So here using bus building algorithm I can calculate the fault current. Okay. So we know very well his bus building algorithm four cases are available. One is reference to new. The next one is existing to new.

[00:12:34]Next one is existing to reference. And another case is existing to existing.

[00:12:42]So based on that we can calculate the Z bus or determine the Z bus. So here in this problem first of all draw the reactance diagram then apply the Z bus building algorithm.

[00:13:01]Okay.

[00:13:08]So this is the reactance diagram for the problem for the given problem. Okay. So here this is so this is generator one transformer T1 this is L1 and L2 and this is transformer T2 this is G2 generator D G G2 the fault occurs in the fourth bus so bus number one bus number two bus number three. So this is bus number four. So fault occurs in the fourth bus.

[00:13:54]Okay. So now find the symmetrical fault current at bus number four.

[00:14:02]So we know the formula for symmetrical fault.

[00:14:06]IF is equal to prefault wtage.

[00:14:12]Okay. divided by Ezinian plus EZF.

[00:14:22]So theian impedance calculated using isbus building algorithm. Okay.

[00:14:34]So very first first write the path for this problem.

[00:14:45]So, so this is actually grounded both the generators are grounded.

[00:14:56]So 0 to from start from here 0 to 1. So the path start from 0 to 1. Next 1 to 2.

[00:15:10]Next 23.

[00:15:15]Next 324.

[00:15:21]And finally 42 0.

[00:15:27]Clear? So 0 is the reference bus. 0 is the reference bus. So bus number one initially it is nothing but new bus. So reference to new reference to new. The second case step number two the bus one is now converted into existing type bus two is new one.

[00:15:52]So here it's nothing but existing to new case.

[00:15:56]The next step 2 3. So second bus is existing. Third bus is new one. So again we can apply existing to new rule. Clear? And the next step 3 to 4 here also third bus number three is existing. Bus number four is new. So again apply existing to new.

[00:16:24]Existing to new and the final four to 0.

[00:16:30]So four is existing bus zero is the reference. So here we have to apply existing to reference existing to reference.

[00:16:44]So now we can solve one by one.

[00:16:53]So first 0 to 1 that is reference to new reference to new. Okay. So in this case.15 is the impedance value between bus number one and zero.

[00:17:23]clear. So now form is bus is equal to J.15 J.15 next so this is step number one this is step number one and step number two step number two is existing to new existing into new clear. So now so an existing to new we can consider 1 22 1 2 bus number 1 2. So bus number 122 the value is 09 09.

[00:18:27]So this part is over.15 clear. Now find the EZ bus. So now the size of the EZ bus is increased to 2 + 2 and the type is existing to new. So the previous value is repeated in the first row and second row.

[00:19:11]So same values repeated in the first row and first column. Similarly second row and the second column the value is added with 09.

[00:19:33]So the previous value is added with the values available between 1 and 2 that is day 0.09.

[00:19:45]So his bus is.15.15.15.

[00:19:53]The next one is 24.

[00:20:01]Next step number three. Step number three is 2 2 3.

[00:20:15]So now bus number two is existing. Three is new one. Again we have to apply existing to new rule.

[00:20:29]existing to new rule.

[00:20:32]Okay. Now it's a value between 2 and 3 05.

[00:21:00]0.05 223.

[00:21:05]So now form the Z bus. So Z bus equal to So size of the Z bus is increased to 2 3 + 3.

[00:21:18]So bus building means we can use the previous bus value. So step by step we can build up the Z bus. So step number two we can get J.152 +2 matrix the same matrix repeated but here the size of the matrix matrix is 3 + 3 So this is the previous result but the size of the matrix here is 3 + 3. So repeat the same column repeated in the third column. The same manner the second row is repeated in the third row.

[00:22:22]So J5 J.24 So here also J.15 J 24 and the last value EZ 33 value is 2.4 4 is added with the element available between 2 and 3 that is J0.05 2 4 plus J 0.05 so 33 value is J 0.29 29 in the same manner we have to apply for uh 324 and 425 also. Okay.

[00:23:34]Next is in step number four we can choose the path 324.

[00:23:41]So this is the path 3 2 4. Okay. Again it's nothing but existing to new role.

[00:23:48]So three is the old bus existing one.

[00:23:52]Fourth bus is introduced in the state.

[00:23:55]That's what it is nothing but a new bus.

[00:23:57]So 3 to four again we are apply existing to new rule. Okay. Existing to new.

[00:24:09]So now the size of the matrix increased to 4 + 4 actually. Another method is also available but the this is the easiest one.

[00:25:11]2.4.

[00:25:13]Yes.

[00:25:18]So in the third row, so in the previous case we are getting 3 + 3 matrix. The same matrix we can implemented here also.

[00:25:27]So, so this is the previous result but in this table the additional row and column is added that is the fourth row and fourth column. Okay. So here uh the fourth row depends the third row value.

[00:25:49]So the third row values repeated in the fourth row. Similarly third column values repeated in the fourth column also.

[00:25:59]Okay. So the same value repeated here.15 here also.24 this is 29.

[00:26:10]Uh then now form B fourth row.15 same third row values repeated.24 24 29 The last Z44 value is J.29 is added with the element available between 3 and 4. So the element available between three and four is 09 09.

[00:26:51]Okay. So, EZ44 value is 38.

[00:26:58]EZ is 44 value is 455. The path is 42 0 is the path. So, this is nothing but existing to reference.

[00:27:11]Existing to reference. So it's similar to existing to new but the only difference is we are applying a chron reduction technique. Clear? So initially size of the matrix is increased to 5 + 5.

[00:27:33]So we can use the same uh previous 4 + 4 matrix and form the fifth row and fifth column using the the additional element that is available between 4 to 0. Okay.

[00:27:48]So 15.

[00:28:07]First write the 4 + 4 matrix. It is obtained in the previous result. Step number four.

[00:28:31]Heat. Heat.

[00:28:38]Heat. Heat.

[00:29:11]So this is the forecast for matrix already obtained in the step number four.

[00:29:18]So say same value repeated in the steps also.

[00:29:43]So it's the last value is obtained 38 right 38.

[00:29:52]So this is a previous uh uh values step number four values forecast for matrix.

[00:29:58]Now form the uh fifth row and fifth column. It exactly similar to existing to new. The only difference is we are applying a chron reduction after forming the 5 + 5 matrix. Right? Okay for understanding and change the font color. So the fourth row repeated in the fifth row.15.24 24 and 29 and 38.

[00:30:34]Okay.

[00:30:35]In the same manner, fifth column 154 29 38 and last is that 55 value is 0.38 is added with the element connected between 4 2 0 that is 15.

[00:31:10]Okay. So that is 55 value is 0 53 53 but this is not original size of the Z bus because it formed based on the virtual node. So eliminate the last row and last column using chron reduction technique. Okay. Now apply So now apply chron reduction technique.

[00:31:48]So purpose of chron reduction is it eliminate the last row and last column.

[00:31:54]Why? Because it created by the last one last column created by the uh virtual node. Okay. So in chron reduction techniques the I value is 1 2 3 dot n minus one because we have to eliminate the last value that is n= 5 fifth row and fifth column and so I not equal to n right similarly j = 1 2 n minus one and j is also not equal to n here. So after applying a chron reduction we are getting the original value of the set bus and size of the bus is 4 + 4 matrix.

[00:32:42]Okay. So what is the chron reduction rule? Is that J new equal to is that I J new equal to is that I J volt is that I J volt minus EZ I N into EZ N J divided EZ N. So this is the crown reduction rule. So as per the rule EZ 1 new equal to EZ12 volt minus EZ 15 into EZ 51 divided by EZ 55. Okay. So is 111 new value is at12 is 15 minus j.15 into j.15 uh divided by last value j 53 So answer for is that 111 new is 111 new equal to J10 75.

[00:34:27]Similarly, EZ12 new equal to EZ12 minus EZ 15 into EZ 52 divided by EZ 55.

[00:34:41]So here E is in E NJ. So I is the for example we can consider starting bus number is I J is N bus number. So is that one two means is at 1 12 means 1 is the starting bus number two is the end bus number. So here is that I n into is that nj n is five. So here I n 5 NJ 52id by is 55. So is 12 new equal to is2 old the same.15 minus is5.15 into is 52 is 52 value is24 is 52 value is24 divided 53 okay so is12 new is so is that 1 to new is 082 07.

[00:35:53]So is at 1 12 new in the same manner is at 13 new is at 13 new equal to is at 13 vol is at 13 volt minus is at 15 into 53 divided by is 55. Okay. So is that 13 is.15 minus J5 into EZ 53 EZ 53 is 29 J.29 divided by J 53.

[00:36:44]So I'll write in the next page. So E is13 E13 new is J 0 6 7 9 6 7 92.

[00:37:01]So this is is that 13 value. So next is at 14 is at 14 new equal to is at 14 old minus is at 15 into is 54 divided by is 55.

[00:37:21]So is that 14 is again.15 minus j.15 into is that 54 is 54 value is that 54 is 38 38 divided by 53.

[00:37:46]Okay.

[00:37:50]So J 04245.

[00:37:57]So this is is that 14 new is at 14 new is at 14 new. So here it's nothing but square matrix right. So is 21 = 1 2 clear. Next is 31 = 1 3. Next is 41 equal to Z4.

[00:38:25]Okay. So next we will find out what is the value of EZ 22 and is 23. So EZ is 22 new EZ 22 new equal to EZ 22 old minus EZ 25 starting bus number is 2. Uh N value is five. So is that 25 in into that NJ is 52 divided by Z 55. Okay. So is that 22 value is 133 the same manner fine.

[00:39:11]EZ 23 EZ 23 new equal to EZ 23 old minus EZ 25 into EZ 53 divided by E is 55. Okay. So what is is 23 volt? Is that 23 volt value?

[00:39:34]Is that 23 value also 24 then is 53 is Z 53 value is 29 okay is 23 is 24 minus is 25 is 25 is also 24 into is 53 is 29 29 divided by 53.

[00:40:08]Okay.

[00:40:11]So answer is J10 86 86 that is EZ 23 new. Okay. So as for the square matrix property EZ 32 equal to is 23. Okay. So no need to find again is 32. And one more thing is is 24 right. Is that 24 is at 24 new equal to is 24 old minus is 25 into is 54 divided by is at 55. So is 24 value is 024 and is at 25 value is also.24 24 into is that 54 value is I think 38 uh 54 is 38 is that 54 value is 38 divided 53 so answer is J 0.067 92.

[00:41:38]So this is at 24 new is at 24 new is at 42 is at 24.

[00:41:53]So the third one is at 31 is available.

[00:41:57]Is that 32 is also available. The only thing is we find out is that 33 is that 43 is 43 is 34 we find out right. So okay in this next step we find out directly is at 33 value and is 34 value is 33 and is 44 value. So 33 volus 35 into 53 divided by 55. Okay. So is 33 new is that 33 new equal to J3 1 3.

[00:42:39]Okay.

[00:42:43]Next is 34 new 34 new equal to Z 34 minus E is Z 35 into EZ 54 divided by EZ 55.

[00:43:01]So Z 34 is also 29 EZ 35.29 29 into 54 38 54 is 38 38 divided 53 3 4 new is that 3 4 new equal to 0 8 20 8.

[00:44:03]So next and finally is at 41 is 42 and is 43 is also available. We find out only is at 44 is at 44 new. Okay. So is 44 old minus 45 into 54 divided by 55. So is that 44 is at 44 is 38. Okay. Is that 54? Is that 54?

[00:44:43]Is that 45? 45 is also 38. Is that 54 also 38.

[00:44:54]So EZ 44 is 38 minus 38² divided by 53 53.

[00:45:18]So this is the final answer 1075.

[00:45:26]Okay. So this is is 44 new.

[00:45:32]So fault occurs in the fourth bus. Okay.

[00:45:37]So we can consider the Z44 value is the EZ tra.

[00:45:43]So ezian is nothing but EZ 44 that is equal to J 1075.

[00:45:58]Okay. So this is the requiredian impedance value. Why? Because the fault occurs in the fourth bus only. Okay. So now find what is the value of if. So in this problem in the given problem there is no fault impedance value. So v teian divided by eian n plus is. So this is the uh formula for symmetrical fault current. So vian the prefault wtage is default one angle 0°. Okay 1 angle 0°.

[00:46:38]Why? because uh generator one and generator 2 in both the case we are getting a 20 KV value right so the prefault value prefault wtage is also same actual voltage and uh base wtage value is also same that's what it's nothing but one angle 0° per unit and is preion is 44 why because fault occurs in the fault occurs in the fourth bus fault occurs in fourth bus fault occurs in fourth bus. Okay. So that we can consider Zion equal to Z 44. So J 1075.

[00:47:22]So here uh they didn't given any fault impedance value. The fault impedance value as per the problem the value is zero. That's nothing but bolted fault.

[00:47:34]The fault impedance value is zero means that is called as bolted fault. Okay. So in this problem uh the set of value is zero. So that one angle 0° divided by J1075.

[00:47:51]So fault current symmetrical.

[00:47:54]So if is equal to 1 angle 0° divided by J1075.

[00:48:02]So if value is minus J 9 3 023.

[00:48:17]So this is a symmetrical fall current value. Symmetrical fall current value current value. And the problem itself we ask the next one is postfault voltage.

[00:48:36]Post fault voltage.

[00:48:42]So postfault wtage how to find post wtage means V suffix I power F equal to 1 angle 0° minus EZ if into F. Okay. So suppose we have to find what is the post voltage at bus number two means V2F equal to 1 angle 0° minus I is bus number. So EZ 2 this F represent bus fault location number fault location number is four. So E is 24 into if okay now check the matrix actually after applying a chron reduction we are getting a 4 + 4 matrix right after applying a chron reduction we are getting 4 + 4 matrix this is the final after chron reduction this is the original is bus matrix okay so this is the original 4 + 4 matrix value. So based on that we can calculate the fault current. So isian value is 1075. Why? Because fault occurs in the fourth bus. So is that 44 is 1075. Fault current is J 9.3023.

[00:50:07]So using the matrix value we can calculate the post wtage. Here Z24 take the matrix value. So second row fourth column is at 24 value is 06792.

[00:50:21]So one angle 0° J 06 792 into the fault current value is minus J 9 3023.

[00:50:41]Okay.

[00:50:43]So bus post fault wtage at bus number two is post fault wtage at bus number two is 1 angle 0° minus 638.

[00:51:04]So final answer 3682 per unit. The same manner we can calculate other two bus wtages that is post fault wtages bus number three we can use V3F.

[00:51:23]V3F is equal to 1 angle 0° minus is that IF. So I I is nothing but three. Fault location number is four. So if the same manner we know very well fault occurs in fourth bus. So V 4F is nothing but zero per unit. Why? Because the fault occurs in the fourth bus only. So one angle 0° minus E is 44 if is nothing but again one angle 0°. So post wtage as bus number four is 0 per unit. So uh this is the procedure for finding a postfault wtage and and fault symmetrical fault current. Okay. So I can solve now uh gate question.

[00:52:12]So uh this is great question paper. So the bus impedance matrix of a three bus system uh they given the matrix 3 + 3 matrix. Here they clearly mentioned the fault impedance value is 0.7 for uni.

[00:52:26]The fault occurs at bus number two. So neglect the prefault loading conditions.

[00:52:32]The voltage at bus one during the fault.

[00:52:35]That means uh question is postfault voltage at bus number one. Okay. So we know the formula for postwaltage uh that is V IF is equal to 1 angle 0° minus E is Z if into if right. So here as per the question we calculate bus postage at bus number one. So v_sub_1 f 1 angle 0° minus ez 1 2. So f represents here is that if this f represent fault location number here the fault occurs at bus number two. So that is 1 12 into if so. So this is the formula for this problem. So first of all find what is the value of if So IF is equal to 1 angle 0° the prefault wtage default it's nothing but one angle 0° and Z terminian plus E is ZF right so one angle 0° divided by EZ 22 why because the fault occurs in second bus right the fault occurs in second bus so the theian impedance is 222 so what is the is 222 value is 222 value is 093 plus fault impedance right so I E22 is 0 93 093 uh that is a fault impedance not fault impedance that is a tin impedance value so now find out what is if so if is one angle 0° uh divided by Z22 value is 093 Right. This is E is tinian plus fault impedance is uh 070 7. Okay. So what is the value of fault current now?

[00:54:45]So answer is minus J 10. So this is the symmetrical fault current. So this is the symmetrical fault current value.

[00:54:55]Okay. Now find voltage at bus number one during the fault. So V_sub_1 F is equal to 1 angle 0° minus E is12. So take the matrix value. So is12. So this is Z12 value. So J 0.061.

[00:55:19]Okay. J 0.061 061 into if into minus J 10.

[00:55:31]Okay. So, so final answer is V1 F is equal to so one angle 0° minus 61. So final answer V1 Final answer is 39 per unit.

[00:55:56]So this is the final answer.

[00:56:08]So 39 per unit.

[00:56:13]So for this problem the bus old voltage at one during the fault that is post fault voltage at bus number one is 39 per unit. So final answer this is one of the gate problem.

[00:56:35]Thank you. Continue in the next video session.