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:

Answer:

Req = 5 + 15||6||12 + 20||80||60 = 5+15||(4+16)||60 = 5+7.5 = 12.5 Ω Ans.

I = V/ Req =40/12.5 = 3.2 A Ans.

3. For the circuit shown bellow, Let VC (0) = 45V. Determine VC (t) and VX (t) at t > 0.

Answer:

Req =12||6 + 8 = 12 Ω

∴ τ = Req. C = 12*1/3 = 4s

∴ VC (t) = VC (0) e-t/ τ

= 15 e - t/4

= 15 e - 0.25t

Ans.

We can use voltage division to get VX (t)

∴ VX (t) = * VC (t) = 1/3* VC (t) = 5 e - 0.25t

Ans.

Reference: Alexandar Sadiku’s book: page 258, example 7.1

Rony Parvej’s EEE Job Solution DPDC-2014, BUET

4. 3-phase line voltages of 2300 volts magnitude are impressed on a balanced wye connected load

which consists of 100Ω resistance per phase in series with 173.2Ω inductive reactance per phase. Find

the line current and the total power taken by the 3-phase load.

Answer:

VL = 2300 V

∴ VΦ = 2300/√3 = 1328 V

ZΦ = 100 + j173.2 = 200∠60º Ω

∴ I Φ = VΦ / ZΦ = 2300 / 200∠60º = 11.5 ∠60º A

∴ IL = 11.5 ∠60º A Ans.

∴ P = 3 VΦ I Φ cos θ = 3*1328*11.5*cos60º = 22908 W = 22.9 KW Ans.

5. Determine the minimum value of IB that will produce saturation. (In saturation VCE ≤ 0.2 V)

Answer:

IC = (10 - 0.2)/4.7 = 2.085 mA

∴ IB = IC / β = 2.085 / 200 = 10.42 μA Ans.

6. Draw the ac small signal equivalent circuit of the following MOSFET configuration.

Rony Parvej’s EEE Job Solution DPDC-2014, BUET

Answer:

Reference: Boylestad: page 429, ch-9, figure: 9.38

7. Calculate the Output VO of the following circuit:

Answer:

The given circuit is a circuit diagram of instrument amplifier. Comparing with the formula of output of

instrument amplifier,

We get,

VO = (12-13)(1+2*5/1)*10/10 = -11V Ans.

Rony Parvej’s EEE Job Solution DPDC-2014, BUET

8. Draw layout diagram of a substation having double bus-bar with single circuit breaker for each

circuit.

9. Show that, the generalized circuit constants of the short transmission line model satisfy the condition

AD - BC =1.

Transmission line can be expressed as a two port network:

Where,

Vs = AVr + BIr - - - - - - - - ①

And Is = CVr + DIr - - - - - - - - ②

A short transmission line can be expressed as,

Here,

Vs = Vr + ZIr - - - - - - - - ③

Is = Ir - - - - - - - - ④

Comparing eq. ① with ③ and ② with ④ we get,

A =1, B=Z, C=0 and D =1.

∴ AD - BC =1*1 – Z*0 = 1- 0 =1. (Showed)

Rony Parvej’s EEE Job Solution DPDC-2014, BUET

10. Write Down the power angle equation for a pure reactance network. Point out a change in which

network element would affect the system satiability.

Factors that influence steady state stability:

(a) Transmission line reactance

(b) Excitation voltage

Factors that influence dynamic stability:

(a) Change in load

(b) Change in turbine speed

Rony Parvej’s EEE Job Solution DPDC-2014, BUET

11. An alternator is supplying a load of 500KVA at 0.6 power factor. If the power factor is raised to

unity, how many more KW can the alternator supply for the same KVA loading?

Answer:

At 0.6 p.f. active power of 500 KVA alternator = 500*0.6 = 300 KW

At unity p.f. active power of 500 KVA alternator = 500*1 = 500 KW

∴ Extra KW that can be supplied by alternator = (500-300) KW = 200 KW Ans.

12. What conditions are required in paralleling two alternators? Draw three-light-bulb method for

checking phase sequence. What else method can check this?

There are five conditions that must be met before the

synchronization process takes place. The alternator must have

Equal line voltage,

Frequency,

Phase sequence,

Phase angle and

Waveform to that of the system to which it is being

synchronized.

[Waveform and phase sequence are fixed by the construction

of the generator and its connections to the system, but voltage,

frequency and phase angle must be controlled each time a

generator is to be connected to a grid.]

Except three-light-bulb method, using a synchroscope method can

also check the phase sequence of an alternator.

Rony Parvej’s EEE Job Solution DPDC-2014, BUET

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

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