231. Four identical alternators each rated for 20 MVA, 11 KV having a subtransient reactance of 16% are working in parallel. The short-circuit level at the bus bars is _______ MVA

1. 1. 5.00

   
   
   

   2. 500

   
   
   

   3. 0.524

   
   
   

   4. 0.52

   
   
   

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1. View Hint View Answer Discuss in Forum

   Short circuit capacity =	Rating of circuit
   	% X

   
   

   =	20	= 500 MVA
   	%4

   Correct Option: B

   Short circuit capacity =	Rating of circuit
   	% X

   
   

   =	20	= 500 MVA
   	%4

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232. A power system network with the capacity at 100 MVA has a source impedance at 10% at a point. The fault at that point is __________ MVA

1. 1. 1000 MVA

   
   
   

   2. 10.7 MVA

   
   
   

   3. 10.1MVA

   
   
   

   4. 10.00 MVA

   
   
   

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1. View Hint View Answer Discuss in Forum

   Fault MVA =	Capacity
   	% of reactance

   
   

   =	100	= 1000 MVA
   	0.10

   Correct Option: A

   Fault MVA =	Capacity
   	% of reactance

   
   

   =	100	= 1000 MVA
   	0.10

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233. The positive sequence current for a L-L fault of a 2 kV system is 1400 A, and corresponding current for a L-L-G fault is 2220 A. The zero sequence impedance of the system is ________ Ω

1. 1. 5.275

   
   
   

   2. 52.75

   
   
   

   3. 5275

   
   
   

   4. 0.5275

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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234. Four identical 100 MVA, 22 kV generators are operating in parallel, as shown above in two busbar sections, interconnected through a current limiting reactor of x p.u. reactance on the generatorbase. Each generator has a reactance of 0.2 p.u. The value of the reactor x to limit a symmetrical short-circuit (a-b-c) current p.u. through the circuit breaker to 1500 MVA is _________ p.u.
     

1. 1. 11.02

   
   
   

   2. 0.12

   
   
   

   3. 0.10

   
   
   

   4. 10

   
   
   

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1. View Hint View Answer Discuss in Forum

   (SC)MVA in p.r. =	1500	= 15
   	100

   
   

   Now, 15 =	1		⇒ 15 =	0.2 + x		= 1.25
   	(0.1 + x)0.1			(0.1 + x)0.1
   	(0.2 + x)

   
   ⇒ 1.5x + 0.15 = 0.2 + x
   ⇒ x = 0.10 p.u.

   Correct Option: C

   (SC)MVA in p.r. =	1500	= 15
   	100

   
   

   Now, 15 =	1		⇒ 15 =	0.2 + x		= 1.25
   	(0.1 + x)0.1			(0.1 + x)0.1
   	(0.2 + x)

   
   ⇒ 1.5x + 0.15 = 0.2 + x
   ⇒ x = 0.10 p.u.

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235. A 50 Hz alternator is rated 500 MVA, 20 kV, with X_d = 1.0 per unit and X_"d = 0.2 per unit. It supplies a purely resistive load of 400 MW at 20 kV. The load is connected directly across the generator terminals when a symmetrical fault occurs at the load terminals. The initial rms current in the generator in per unit is __________

1. 1. 3.022

   
   
   

   2. 3.22

   
   
   

   3. 53.26

   
   
   

   4. 542.5

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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