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UPRVUNL’21 EE: Daily Practices Quiz 12-July-2021

Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 06
Time: 08 min.

Q1. The flux is maximum in the following part of a dc motor?
(a) Under trailing pole tip
(b) Under leading pole tip
(c) Pole core
(d) Under the interpole

Q2. Which testing method is performed to determine no load losses in DC motors?
(a) Field test
(b) Brake test
(c) Running down test
(d) Swinburne’s test

Q3. A 6-pole, 50 Hz, 1-phase induction motor runs at a speed of 900 r.p.m. The frequency of currents in the cage rotor will be
(a) 5 Hz, 50 Hz
(b) 5 Hz, 55 Hz
(c) 5 Hz, 95 Hz
(d) 55 Hz, 95 Hz

Q4. A series resonant circuit is tuned to 10 MHz and provides 3-dB bandwidth of 100 kHz. The quality factor Q of the circuit is
(a) 1
(b) 10
(c) 30
(d) 100

Q5. Fermi energy level for P-type extrinsic semiconductor lies….
(a) Close to valence band
(b) Close to conduction band
(c) At the middle of the band gap
(d) None of the above

Q6. If voltage is increased by n times, the size of the conductor would
(a) increase by n times
(b) reduce by 1/n times
(c) increase by n² times
(d) reduce by 1/n² times

S1. Ans.(b)
Sol. In a DC Generator the armature reaction strengthens the flux at a trialing tips and weakens the flux at a leading pole tips, whereas the armature reaction in a DC motor creates the opposite effects.

S2. Ans.(d)
Swinburne’s Test: – for large shunt and compound wound motor testing
Direct loading test: – For small motor.
So, basically Swinburne’s test is an indirect method of testing, in which DC machine is run as motor at no load and rated speed and voltage. By this test, losses, efficiency and temperature rise etc. are Calculated.

S3. Ans.(c)
Sol. Synchronous speed for forward rotating flux, Ns = 120 f/P = 120 × 50/6 = 1000 r.p.m.
Therefore, slip for forward rotating flux is
s_f=s=(N_s-N)/N_s =(1000-900)/1000=0.1
∴ Rotor current frequency for sf is f’ = sf × f = 0.1 × 50 = 5 Hz.
Slip for backward rotating flux s_b =(2 – s) =(2 – 0.1) = 1.9. Therefore, rotor current frequency, f” = (2 – s) × f = 1.9 × 50 = 95 Hz.

S4. Ans.(d)
Sol. for resonant circuit, Q=f_0/BW=(10×10^6)/(100×10^3 )=100

S5. Ans.(a)
Sol. Fermi energy level (E_F) for P- type extrinsic semiconductors lie close to valence band and for N-type, it is close to conduction band.

uprvunl je electrical quiz |_30.1

S6. Ans.(d)
Sol. Area of cross-section of conductor ∝ 1/V².


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