171. A three-phase, 33 kV oil circuit breaker is rated 1200 A, 2000 M VA, 3s. The symmetrical breaking current is

1. 1. 1200 A

   
   
   

   2. 3600 A

   
   
   

   3. 35 kA

   
   
   

   4. 104.8 kA

   
   
   

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1. View Hint View Answer Discuss in Forum

   Symmetrical breaking current =		MVA			kA
   		√3kV

   
   

   =	2000
   	√3 × 33

   
   = 34.99 kA
   ≈ 35 kA

   Correct Option: C

   Symmetrical breaking current =		MVA			kA
   		√3kV

   
   

   =	2000
   	√3 × 33

   
   = 34.99 kA
   ≈ 35 kA

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172. Consider two buses connected by an impedance of (0 + j5)Ω. The bus 1 voltage is 100 ∠30° V, and bus 2 voltage is 100 ∠0° V. The real and reactive power supplied by bus 1, respectively are

1. 1. 1000 W, 268 VAr

   
   
   

   2. – 1000 W, – 134VAr

   
   
   

   3. 276.9 W, – 56.7 VAr

   
   
   

   4. – 276.9 W, 56.7 VAr

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   P + jQ = VI* = 100 ∠30°		100∠30° - 100∠0°
   		5j

   
   = 100 ∠30° {20∠-60 - 20∠- 90°}*
   =100 ∠30° {(10 - j17.32) + 20j}*
   =100 ∠30° . 10.35 ∠- 15°
   = 1035 ∠15° = P + jQ
   ∴ P = 1035 cos 15 = 1000W
   and Q = 1035 sin 15 = 268 VAr.

   Correct Option: A

   
   

   P + jQ = VI* = 100 ∠30°		100∠30° - 100∠0°
   		5j

   
   = 100 ∠30° {20∠-60 - 20∠- 90°}*
   =100 ∠30° {(10 - j17.32) + 20j}*
   =100 ∠30° . 10.35 ∠- 15°
   = 1035 ∠15° = P + jQ
   ∴ P = 1035 cos 15 = 1000W
   and Q = 1035 sin 15 = 268 VAr.

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173. Consider a step voltage wave of magnitude 1 pu travelling along a lossless transmission line that terminates in a reactor. The voltage magnitude across the reactor at the instant the travelling wave reaches the reactor is
     

1. 1. – 1 pu

   
   
   

   2. 1 pu

   
   
   

   3. 2 pu

   
   
   

   4. 3 pu

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   E(s) = 1/S [step voltage wave having magnitude 1 p.u.]
   Voltage across the inducter is,
   

   V(s) =	E(s)	=	E(s)	LS
   	Zc + LS		Zc + LS

   
   

   V(s) =	1
   	S +	Zc
   		L

   
   Taking inverse Laplace transform, we get v(t)
   v(t) = e^(-z_c/L)t
   At
   t = 0
   v(t) = 1 p.u.

   Correct Option: A

   
   E(s) = 1/S [step voltage wave having magnitude 1 p.u.]
   Voltage across the inducter is,
   

   V(s) =	E(s)	=	E(s)	LS
   	Zc + LS		Zc + LS

   
   

   V(s) =	1
   	S +	Zc
   		L

   
   Taking inverse Laplace transform, we get v(t)
   v(t) = e^(-z_c/L)t
   At
   t = 0
   v(t) = 1 p.u.

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174. A balanced three-phase voltage is applied to a star-connected induction motor, the phase to neutral voltage being V. The stator resistance, rotor resistance referred to the stator, stator leakage reactance, rotor leakage reactance refer red to the stator, and the magnet izing reactance are denoted by r_s, r_r, x_s, x_r and X_m, respectively. The magnitude of the starting current of the motor is given by

1. 1. V
      √(rs + r'r)² + (Xs + X'r)²

   
   
   

   2. V
      √rs² + (Xs + Xm)²

   
   
   

   3. V
      √(rs + rr)² + (Xm + Xr)²

   
   
   

   4. V
      √rs² + (Xm + Xr)²

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   ∴	V
   	√(rs + r'r)² + (Xs + X'r)²

   Correct Option: A

   
   

   ∴	V
   	√(rs + r'r)² + (Xs + X'r)²

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175. Power is transferred from system A to system B by an HVDC link as shown in the figure given below. If the voltages V_AB and V_CD are as indicated in the figure, and I > 0, then
     

1. 1. VAB < 0, VCD < 0, VAB > VCD

   
   
   

   2. VAB > 0, VCD > 0, VAB < VCD

   
   
   

   3. VAB > 0, VCD > 0, VAB > VCD

   
   
   

   4. VAB > 0, VCD < 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   V_AB > 0 since rectifier output
   V_CD > 0 inverter output
   For I > 0, V_AB > V_CD,
   hence only current will flow in the given direction.

   Correct Option: C

   V_AB > 0 since rectifier output
   V_CD > 0 inverter output
   For I > 0, V_AB > V_CD,
   hence only current will flow in the given direction.

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