Engineering Jobs   »   Quiz: Electrical Engineering 23 July 2020

Quiz: Electrical Engineering 23 July 2020

Quiz: Electrical Engineering
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. A 4 KVA, 400/200 V single-phase transformer has resistance of 0.02 pu and reactance of 0.06 pu. Its actual resistance and reactance referred to HV side are respectively:
(a)0.2 ohm and 0.6 ohm
(b)0.8 ohm and 2.4 ohm
(c)0.08 ohm and 0.24 ohm
(d)2 ohm and 6 ohms

Q2. A transformer has a percentage resistance of 2% and percentage reactance of 4%. What are its regulations at power factor 0.8 lagging and 0.8 leading, respectively?
(a) 4% and – 0.8%
(b) 3.2% and – 1.6%
(c) 1.6% and – 3.2 %
(d) 4.8% and – 0.6%

Q3. If the iron core of a transformer is replaced by an air core, then the hysteresis losses in the transformer will
(a) increase
(b) decrease
(c) remain unchanged
(d) become zero

Q4. The use of higher flux density in the transformer design
(a) Reduces the weight per kVA
(b) Increases the weight per kVA
(c) Has no relation with the weight of transformer
(d) Increases the weight per kW

Q5. Which interrupt has the highest priority in 8085?
(c)RST 6.5
(d)RST 7.5

Q6. An induction motor has a rotor resistance of 0.002Ω/phase. If the resistance increased to 0.004Ω/phase then the maximum torque will
(a) Reduced to half
(b) increased by 100%
(c) Increased by 200%
(d) Remain unaltered

Q7. Cogging of induction motor occurs at
(a)High voltage and when stator and rotor poles are equal
(b)Low voltage and when stator and rotor poles are equal
(c)High voltage and when stator and rotor poles are not equal
(d)Low voltage and when stator and rotor poles are not equal

Q8. In star connected system the phase angle difference between line and phase voltage is:
(a) 30°
(b) 120°
(c) 60°
(d) 90°

Q9. The phase-lead compensation is used to
(a)Increase rise time and decrease overshoot
(b)Decrease both rise time and overshoot
(c)Increase both rise time and overshoot
(d)Decrease rise time and increase overshoot.

Q10. An OP-AMP has a common mode gain of 0.01 and a differential gain of 10^5. Its common mode rejection ratio will be


S1. Ans.(b)
Sol. per unit (pu) quantity will be same on both side because the per-unit impedance is the same for the equivalent circuit of the transformer whether computed from the primary or secondary as long as the voltage bases on the two sides are selected in the ratio of transformation.
On HV side: Z_base=(VH^2)/S=(400)^2/4000=40 ohm
And Zpu=0.02+j 0.06
∴Zactual=Zbase×Zpu=0.8+j 2.4 ohm

S2. Ans.(a)
Sol. Per unit regulation:

%regulation, lagging = 2*0.8+4*0.6= 4%
%regulation, leading = 2*0.8-4*0.6= -0.8%

S3. Ans.(d)
Sol. Hysteresis is the property exhibited by ferrous materials, particularly those having iron in its composition. So naturally air cored transformers will not have any hysteresis losses.

S4. Ans.(a)
Sol. B=Φ/A. for same Φ, A is less. so, weight will reduce.

S5. Ans.(b)
Sol. Priority order of interrupt in 8085:
TRAP > RST 7.5 > RST 6.5 > RST 5.5 > INTR
So, in 8085 TRAP has highest priority.

S6. Ans.(d)
Sol. Maximum torque in an induction motor, T_MAX ∝1/2X.
the maximum torque is independent of rotor resistance and is inversely proportional to rotor reactance. So, if we increase the rotor resistance the maximum torque will be unaltered.

S7. Ans.(b)
Sol. Starting Torque ∝ V². When the supply voltage is low, the starting torque is low and when the poles on stator and rotor are equal, due to locking tendency motor caused by the slot harmonics reduced/failed to start.

S8. Ans.(a)
Sol. Note: – for balanced star-connected system V_L=√3 V_ph and line voltage leads the phase voltage by 30°.

S9. Ans.(b)
Sol. Phase-lead compensator is used to decrease both rise time and overshoot.

S10. Ans.(d)
Sol. CMRR=Ad/Ac =10^5/0.01=10^7

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