-
259.
Statement for Linked Answer Questions :
Cayley –Hamilton Theorem states that a square matrix satisfies its own characteristic equation. Consider a matrix
[1] A satisfies the relation [2 marks]
(A) A + 3I + 2A-1 = 0
(B) A2 + 2A + 2I = 0
(C) (A+I) (A+2I) = 0
(D) exp (A) = 0[2] A9 equals [2 marks]
(A) 511 A + 510 I
(B) 309 A + 104 I
(C) 154 A + 155 I
(D) exp (9A)asked in Electrical Engineering, 2007
View Comments [0 Reply]
-
260.
Statement for Linked Answer Questions :
An inductor designed with 400 turns coil wound on an iron core of 16 cm2 cross sectional area and with a cut of an air gap length of 1mm. The coil is connected to a 230 V, 50 Hz ac supply. Neglect coil resistance, core loss, iron reluctance and leakage inductance. (μo=4π × 10-7 H/m).[1] The current in the inductor is [2 marks]
(A) 18.08 A
(B) 9.04 A
(C) 4.56 A
(D) 2.28 A[2] The average force on the core to reduce the air gap will be [2 marks]
(A) 832.29 N
(B) 1666.22 N
(C) 3332.47 N
(D) 6664.84 Nasked in Electrical Engineering, 2007
View Comments [0 Reply]
-
261.
A1: 1 Pulse transformer (PT) is used to trigger the SCR in the adjacent figure. The SCR is rated at 1.5kV, 250 A with IL=250 mA, IH = 150 mA, and IGmax=150 mA, IGmin =100 mA. The SCR is connected to an inductive load, where L=150 mH in series with a small resistance and the supply voltage is 200 Vdc. The forward drops of all transistors/ diodes and gate- cathode junction during ON state are 1.0 V.
[1] The resistance R should be [2 marks]
(A) 4.7 k Ω
(B) 470 Ω
(C) 47 Ω
(D) 4.7Ω[2] The minimum approximate volt-second rating of the pulse transformer suitable for triggering the SCR should be: (volt-second rating is the maximum of product of the voltage and the width of the pulse that may be applied)
[2 marks]
(A) 2000 μV-s
(B) 200 μV-s
(C) 20 μV-s
(D) 2.0 μV-sasked in Electrical Engineering, 2007
View Comments [0 Reply]
-
262.
A three phase squirrel cage induction motor has a starting current of seven times the full load current and full load slip of 5%.
[1] If an autotransformer is used for reduced voltage starting to provide 1.5 per unit starting torque, the autotransformer ratio (%) should be
[2 marks]
(A) 57.77%
(B) 72.56%
(C) 78.25%
(D) 81.33%[2] If a star-delta starter is used to start this induction motor, the per unit starting torque will be [2 marks]
(A) 0.607
(B) 0.816
(C) 1.225
(D) 1.616[3] If a starting torque of 0.5 per unit is required then the per unit starting current should be [2 marks]
(A) 4.65
(B) 3.75
(C) 3.16
(D) 2.13asked in Electrical Engineering, 2007
View Comments [0 Reply]
-
263.
-
264.
