Rony Parvej’s EEE Job Solution DPDC-2014, BUET Solution of Recruitment Test Question of Dhaka Power Distribution Company (DPDC) Solved by: Rony Parvej (IUT, EEE’07) Post: Assistant Engineer (Electrical) Time: 90 minutes Full Marks: 100 Exam Date: 07.11.2014 Venue: BUET 2. Determine R eq and I in the circuit shown bellow: Answer: R eq = 5 + 15||{6||12 + 20||80}||60 = 5+15||(4+16)||60 = 5+7.5 = 12.5 Ω Ans. I = V/ R eq =40/12.5 = 3.2 A Ans. 3. For the circuit shown bellow, Let V C (0) = 45V. Determine V C (t) and V X (t) at t > 0. Answer: R eq =12||6 + 8 = 12 Ω ∴ τ = R eq . C = 12*1/3 = 4s ∴ V C (t) = V C (0) e -t/ τ = 15 e - t/4 = 15 e - 0.25t Ans. We can use voltage division to get V X (t) ∴ V X (t) = * V C (t) = 1/3* V C (t) = 5 e - 0.25t Ans. Reference: Alexandar Sadiku’s book: page 258, example 7.1
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DPDC (07.11.2014) Recruitment Test Solution by Rony Parvej (Departmental Written)
DPDC (07.11.2014) Recruitment Test Solution by Rony Parvej (Departmental Written)
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Rony Parvej’s EEE Job Solution DPDC-2014, BUET
Solution of Recruitment Test Question of Dhaka Power Distribution Company (DPDC) Solved by: Rony Parvej (IUT, EEE’07)
Post: Assistant Engineer (Electrical)
Time: 90 minutes Full Marks: 100
Exam Date: 07.11.2014 Venue: BUET
2. Determine Req and I in the circuit shown bellow:
13. Draw an auto-transformer starter for a 3-phase induction motor and mention the starting sequence.
The starting sequence is:-
The Star switch is closed
The Start switch is closed to energize the autotransformer
The motor is connected at a selected reduced voltage tap on the autotransformer and
starts to turn and accelerate
After a predetermined period the Star switch will open
After a mili-second delay the Run switch will close, connecting full line voltage to the
motor
The Start switch will then open and the motor will be at operational speed
14. A PCM-TDM system multiplexes 10 band limited voice channel (300-3400 Hz) and uses a 256
level quantizer. If the signal is sampled at a rate 17 % higher than Nyquist rate, then what will be the
maximum energy bandwidth of the transmission channel?
Here,
Maximum frequency of the message signal, fm = 3400 Hz.
∴ Nyquist frequency of the signal, fNQ = 2 fm = 2* 3400 Hz =6.8 KHz
∴ Sampling frequency of the signal, fS = 17 % (=17.65%) higher than Nyquist rate
= 1.1765*6.8 KHz = 8KHz
Quantization Level, L = 256 = 28
∴ No. of bits in the code, n=8
∴ Maximum energy bandwidth of one channel = n fS = 8*8 KHz = 64 KHz.
∴ Maximum energy bandwidth of 10 channel = 10* 64 KHz =640 KHz Ans.
Rony Parvej’s EEE Job Solution DPDC-2014, BUET
15. The spectrum of a modulating signal is shown in the figure. Draw the spectrum of DSB-SC,
SSB+C, and VSB modulated signals for this modulating signal assuming a carrier signal of
C (t) = AC Cos 2π fC’ t
Answer:
------------------------------ 0 ----------------------------------- এটা ১০০% সঠিক সরান না। আভায কাছে যমটা সঠিক ভছন হছেছে তা ফরা মাে। যকান বর যছর নীছেয যম যকান একঠি এছেছস
জানাছনায অনছযাধ যইছরা।
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