Ans: D
N = Ns (1-S) 950 = 120 x 50 (1-S)/6 S = 0.05
Q.86 Reduction in the capacitance of a capacitor-start motor, results in reduced
(A) Noise. (B) Speed. (C) Starting torque. (D) Armature reaction.
Ans: C
Reduction in the capacitance reduces starting voltage, which results in reduced starting torque.
Q.87 Regenerative braking (A) Can be used for stopping a motor. (B) Cannot be easily applied to DC series motors. (C) Can be easily applied to DC shunt motors (D) Cannot be used when motor load has overhauling characteristics.
Ans: B Because reversal of Ia would also mean reversal of field and hence of Eb
Q.88 At present level of technology, which of the following method of generating electric power from sea is most advantageous? (A) Tidal power. (B) Ocean thermal energy conversion (C) Ocean currents. (D) Wave power.
Ans: A At present level of technology, tidal power for generating electric power from sea is most advantageous because of constant availability of tidal power.
Q.89 If the field circuits of an unloaded salient pole synchronous motor gets suddenly open circuited, then (A) The motor stops. (B) It continues to run at the same speed. (C) Its runs at the slower speed. (D) It runs at a very high speed.
Ans: B The motor continues to run at the same speed because synchronous motor speed does not depend upon load, Nα f.
Q.90 Electric resistance seam welding uses __________ electrodes.
(A) Pointed (B) Disc. (C) Flat (D) Domed
Ans: B Disc type electrodes are used for electric resistance seam welding.
Q.91 For LV applications (below 1 kV), ______________ cables are used.
(A) Paper insulated. (B) Plastic. (C) Single core cables. (D) Oil filled.
Ans: C For low voltage applications single core cables are suitable.
Q.92 No load current in a transformer:
(A) lags the applied voltage by 90° (B) lags the applied voltage by somewhat less than 90°
(C) leads the applied voltage by 90° (D) leads the applied voltage by somewhat less than 90°
Ans: B
The primary input current under no load conditions has to supply (i) iron losses in the core i.e hysteresis loss and eddy current loss (ii) a very small amount of Cu loss in the primary (there being no Cu loss in secondary as it is open)
Q.93 A transformer operates most efficiently at 3/4th full load. Its iron (PI) and copper loss (PCu) are related as:
(A) 169 =CuI PP (B) 43 =CuI PP (C) 34 =CuI PP (D) 916 =CuI PP
Ans: D
If PCu is the Cu loss at full load, its value at 75% of full load is PCu x (0.75)2 = 9/16 PCu
At maximum efficiency, it equals the iron loss PI which remains constant through out. Hence max. efficiency at
PI = 9/16 PCu Or PI / PCu = 9/16
Q.94 In a salient pole synchronous machine (usual symbols are used):
(A) qd xx > (B) qd xx = (C) qd xx < (D) 0 =qx
Ans: C Since reluctance on the q axis is higher, owing to the larger air gap, hence xq < xd
Q.95 The armature of a dc machine is laminated to reduce:
(A) Eddy current loss (B) Hysteresis loss (C) copper losses (D) friction and windage losses
Ans: A
Thinner the laminations, greater is the resistance offered to the induced e.m.f., smaller the current and hence lesser the I2R loss in the core.
Q.96 The resistance representing mechanical output in the equivalent circuit of an induction motor as seen from the stator is:
(A) −112 s r' (B) s r' 2 (C) −112 2 s r (D) s r2
Ans: A Mechanical Power developed by the rotor (Pm) or gross power developed by rotor (Pg) = rotor input –rotor Cu losses = (3I/2 R2/ / S) -(3I/2 R2/ ) = 3I/2 R2/ (1/ S -1)
Q.97 A single phase Hysteresis motor
(A) can run at synchronous speed only
(B) can run at sub synchronous speed only
(C) can run at synchronous and super synchronous speed
(D) can run at synchronous and sub synchronous speed
Ans: A
The rotor revolves synchronously because the rotor poles magnetically lock up with the revolving stator poles of opposite polarity
Q. 98 The temperature of resistance furnaces can be controlled by changing the:
(A) applied voltage (B) number of heating elements
(C) circuit configuration (D) All of the above
Ans: D
Temperature of resistance furnaces can be controlled by changing either applied voltage or by number of heating elements or by circuit configuration.
Q.99 The line trap unit employed in carrier current relaying:
(A) offers high impedance to 50 Hz power frequency signal
(B) offers high impedance to carrier frequency signal
(C) offers low impedance to carrier frequency signal
(D) Both (A) & (C)
Ans: B
The line trap unit employed in carrier current relaying offers high impedance to carrier frequency signal.
Because carrier frequency range is 35 km – 500 kHz
XL = 2Π f l Where f increases XL will also increases
Q.100 For a line voltage V and regulation of a transmission line R (A) V ∝R (B) V 1 ∝R (C) 2 V∝R (D) 2 V 1 ∝R
Ans: B
R α 1/V
Regulation = (V0 – VL ) / V0 , if VL is high the (V0 – VL ) will be low.
Therefore R α 1/V
N = Ns (1-S) 950 = 120 x 50 (1-S)/6 S = 0.05
Q.86 Reduction in the capacitance of a capacitor-start motor, results in reduced
(A) Noise. (B) Speed. (C) Starting torque. (D) Armature reaction.
Ans: C
Reduction in the capacitance reduces starting voltage, which results in reduced starting torque.
Q.87 Regenerative braking (A) Can be used for stopping a motor. (B) Cannot be easily applied to DC series motors. (C) Can be easily applied to DC shunt motors (D) Cannot be used when motor load has overhauling characteristics.
Ans: B Because reversal of Ia would also mean reversal of field and hence of Eb
Q.88 At present level of technology, which of the following method of generating electric power from sea is most advantageous? (A) Tidal power. (B) Ocean thermal energy conversion (C) Ocean currents. (D) Wave power.
Ans: A At present level of technology, tidal power for generating electric power from sea is most advantageous because of constant availability of tidal power.
Q.89 If the field circuits of an unloaded salient pole synchronous motor gets suddenly open circuited, then (A) The motor stops. (B) It continues to run at the same speed. (C) Its runs at the slower speed. (D) It runs at a very high speed.
Ans: B The motor continues to run at the same speed because synchronous motor speed does not depend upon load, Nα f.
Q.90 Electric resistance seam welding uses __________ electrodes.
(A) Pointed (B) Disc. (C) Flat (D) Domed
Ans: B Disc type electrodes are used for electric resistance seam welding.
Q.91 For LV applications (below 1 kV), ______________ cables are used.
(A) Paper insulated. (B) Plastic. (C) Single core cables. (D) Oil filled.
Ans: C For low voltage applications single core cables are suitable.
Q.92 No load current in a transformer:
(A) lags the applied voltage by 90° (B) lags the applied voltage by somewhat less than 90°
(C) leads the applied voltage by 90° (D) leads the applied voltage by somewhat less than 90°
Ans: B
The primary input current under no load conditions has to supply (i) iron losses in the core i.e hysteresis loss and eddy current loss (ii) a very small amount of Cu loss in the primary (there being no Cu loss in secondary as it is open)
Q.93 A transformer operates most efficiently at 3/4th full load. Its iron (PI) and copper loss (PCu) are related as:
(A) 169 =CuI PP (B) 43 =CuI PP (C) 34 =CuI PP (D) 916 =CuI PP
Ans: D
If PCu is the Cu loss at full load, its value at 75% of full load is PCu x (0.75)2 = 9/16 PCu
At maximum efficiency, it equals the iron loss PI which remains constant through out. Hence max. efficiency at
PI = 9/16 PCu Or PI / PCu = 9/16
Q.94 In a salient pole synchronous machine (usual symbols are used):
(A) qd xx > (B) qd xx = (C) qd xx < (D) 0 =qx
Ans: C Since reluctance on the q axis is higher, owing to the larger air gap, hence xq < xd
Q.95 The armature of a dc machine is laminated to reduce:
(A) Eddy current loss (B) Hysteresis loss (C) copper losses (D) friction and windage losses
Ans: A
Thinner the laminations, greater is the resistance offered to the induced e.m.f., smaller the current and hence lesser the I2R loss in the core.
Q.96 The resistance representing mechanical output in the equivalent circuit of an induction motor as seen from the stator is:
(A) −112 s r' (B) s r' 2 (C) −112 2 s r (D) s r2
Ans: A Mechanical Power developed by the rotor (Pm) or gross power developed by rotor (Pg) = rotor input –rotor Cu losses = (3I/2 R2/ / S) -(3I/2 R2/ ) = 3I/2 R2/ (1/ S -1)
Q.97 A single phase Hysteresis motor
(A) can run at synchronous speed only
(B) can run at sub synchronous speed only
(C) can run at synchronous and super synchronous speed
(D) can run at synchronous and sub synchronous speed
Ans: A
The rotor revolves synchronously because the rotor poles magnetically lock up with the revolving stator poles of opposite polarity
Q. 98 The temperature of resistance furnaces can be controlled by changing the:
(A) applied voltage (B) number of heating elements
(C) circuit configuration (D) All of the above
Ans: D
Temperature of resistance furnaces can be controlled by changing either applied voltage or by number of heating elements or by circuit configuration.
Q.99 The line trap unit employed in carrier current relaying:
(A) offers high impedance to 50 Hz power frequency signal
(B) offers high impedance to carrier frequency signal
(C) offers low impedance to carrier frequency signal
(D) Both (A) & (C)
Ans: B
The line trap unit employed in carrier current relaying offers high impedance to carrier frequency signal.
Because carrier frequency range is 35 km – 500 kHz
XL = 2Π f l Where f increases XL will also increases
Q.100 For a line voltage V and regulation of a transmission line R (A) V ∝R (B) V 1 ∝R (C) 2 V∝R (D) 2 V 1 ∝R
Ans: B
R α 1/V
Regulation = (V0 – VL ) / V0 , if VL is high the (V0 – VL ) will be low.
Therefore R α 1/V
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