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UPRVUNL’21 EE: Daily Practices Quiz 09-Sep-2021

UPRVUNL JE Recruitment 2021: 

A total of 196 vacancies were released by the UPRVUNL Recruitment 2021. The selection process will be based on the written examination which will be divided into two parts – Part I and Part II. Part I will have 150 objective-type questions from the syllabus for the Diploma Engineering in the relevant branch. While Part II will have 50 objective-type questions comprising General Knowledge, General Hindi, and Reasoning.

UPRVUNL’21 EE:  Daily Practices Quiz 09-Sep-2021

Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 06
Time: 06 min.

Q1. A meter has a square law scale. For 2 A current the deflection in 90°. For a deflection of 45°, the current is
(a) 1 A
(b) 1.5 A
(c) 1.414 A
(d) 1.61 A

Q2. Which of the following is involved in the operation of a JFET?
(a) Flow of minority carriers alone
(b) Flow of majority carriers alone
(c) Flow of both majority and minority carriers
(d) Use of magnetic field

Q3. In which mode the JFET can operate?
(a) Depletion mode only
(b) Enhancement mode only
(c) Saturation mode only
(d) Noise mode only

Q4. What is the use of higher flux density in the transformer design?
(a) For increasing the weight/KVA
(b) For decreasing the weight/KVA
(c) For reducing iron losses
(d) For improving insulation

Q5. A CRO uses:
(a) Electromagnetic focusing
(b) Electrostatic focusing
(c) Both focusing techniques
(d) No focusing technique

Q6. A series R-L-C circuit has a resonant frequency f_0 with R=1 Ω,L=1 H and C=1 F. If the component’s values are doubled, the new resonant frequency will be
(a) f0
(b) 2f0
(c) f0/2
(d) unaltered

S1. Ans.(c)
Sol. For square scale: Ө α I^2
∴(I_2^2)/(I_1^2 )=Ө_2/Ө_1
⇒I_2=√(Ө_2/Ө_1 )×I_1=(45/90)^(1/2)×2=√2=1.414 A

S2. Ans.(b)
Sol. only majority carriers are involved in creations of current in JFET, it is a unipolar device. The input impedance of a JFET is very high.

S3. Ans.(a)
Sol. JFET only works in the depletion mode, whereas MOSFETs have depletion mode and enhancement mode.

S4. Ans.(b)
Sol. The value of flux density in the core determines the core area.
High value of flux density gives smaller core area, so saving in iron cost. Also, small core area provides reduced mean turn of winding which gives reduction in copper cost. But
higher flux density increases iron losses resulting high temperature rise.

S5. Ans.(b)
Sol. A CRO uses electrostatic focusing.

S6. Ans.(c)
Sol. resonant frequency(f)=1/√LC
If components are doubled then f^’=1/√(2L×2C)=f/2

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