226. If the fault current is 2000 A, the relay setting is 50% and CT ratio is 400: 5, then plug setting multiplier will be _________ A

1. 1. 10

   
   
   

   2. .010

   
   
   

   3. 1.0

   
   
   

   4. .001

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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227. A small generating station has a busbar divided into three sections each section is connected to a tie bar with reactor each rated at 5 mVA, 0.1 p.u. reactance. A generator of 8mVA, rating and 0.15 p.u. reactance is connected to each section of the busbar. The short circuit capacit y of the breaker, if a 3 – φ fault takes place on one of the sections of the busbar, will be _________ mVA

1. 1. 7.873 MVA

   
   
   

   2. 78.73 MVA

   
   
   

   3. 787.3 MVA

   
   
   

   4. 7873 MVA

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the base MVA be 8MVA,
   Per unit reactance of the generator is 0.15 p.u., and per unit resistance of the reactor
   

   = 0.1 ×	8	= 0.16 p.u.
   	5

   
   
   Equivalent impedance,
   

   Zeq =	j0.15 × j0.315	= j0.10161
   	j0.465

   
   Short circuit capacity
   

   =	Base MVA	=	8	= 78.73 MVA
   	j0.10161

   Correct Option: B

   Let the base MVA be 8MVA,
   Per unit reactance of the generator is 0.15 p.u., and per unit resistance of the reactor
   

   = 0.1 ×	8	= 0.16 p.u.
   	5

   
   
   Equivalent impedance,
   

   Zeq =	j0.15 × j0.315	= j0.10161
   	j0.465

   
   Short circuit capacity
   

   =	Base MVA	=	8	= 78.73 MVA
   	j0.10161

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228. The line currents in ampere in phases a, b and c respectively are (400 + j100), (75 – j600) and (– 300 + j500) referred to the same reference vector. The negative sequence components of currents will be _______ Amp.

1. 1. 151.9

   
   
   

   2. 159.9

   
   
   

   3. 148.9

   
   
   

   4. 149.9

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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229. When a line-to-ground fault occurs, the current in the phase is 100 A. The zero sequence current in this case will be _______ A

1. 1. 1.4996 |202.5°

   
   
   

   2. 149.96 |200°

   
   
   

   3. 149.96 |202.5°

   
   
   

   4. 14900 |202.5°

   
   
   

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1. View Hint View Answer Discuss in Forum

   For a L – G fault,
   I_a0 = I_a1 = I_a2 and I_a = 3 I_a1
   

   ∴ Ia0 =	Ia	=	100	= 33.3 A 212
   	3		3

   
   I_a =(400 + j 100), I_b = (75 – j 600), I_c = (– 300 + j 500)
   I_a2 = 1/3 (I_a + α² I_b + αI_c)
   

   =	1	[(400 + j100) + (75 - j 600) |240° + (-300 + j 500)|120°]
   	3

   
   

   =	1	[400 + j100 + (75 - j 600) + (-300 + j 500)(-0.5 + j0.86)]
   	3

   
   

   =	1	[(400 + j100) + -553.5 + j235.5 - 280 - j 508]
   	3

   
   

   =	1	[ - 415.5 - j 172.5]
   	3

   
   = 149.96 |202.5°

   Correct Option: C

   For a L – G fault,
   I_a0 = I_a1 = I_a2 and I_a = 3 I_a1
   

   ∴ Ia0 =	Ia	=	100	= 33.3 A 212
   	3		3

   
   I_a =(400 + j 100), I_b = (75 – j 600), I_c = (– 300 + j 500)
   I_a2 = 1/3 (I_a + α² I_b + αI_c)
   

   =	1	[(400 + j100) + (75 - j 600) |240° + (-300 + j 500)|120°]
   	3

   
   

   =	1	[400 + j100 + (75 - j 600) + (-300 + j 500)(-0.5 + j0.86)]
   	3

   
   

   =	1	[(400 + j100) + -553.5 + j235.5 - 280 - j 508]
   	3

   
   

   =	1	[ - 415.5 - j 172.5]
   	3

   
   = 149.96 |202.5°

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230. In a power system with negligible resistance, the fault current at a point is 8.00 p.u. The series reactance to be included at the fault point to limit the short-circuit current to 5.00 p.u. is ______p.u.

1. 1. 3000

   
   
   

   2. 3.000

   
   
   

   3. 0.300

   
   
   

   4. 30.00

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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