MAHADISCOM Sub Engineer Question Papers

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You are asking for the Maharashtra State Electricity Distribution Company ( MAHADISCOM) Sub Engineer Question Papers. Here I am uploading a file that contains the MAHADISCOM Sub Engineer Question Papers. You can download this from here. This is as follows:
MAHADISCOM Sub Engineer Question Papers





//mahadiscom.in/Professional-Exam-72Part-I.pdf

Here I am also uploading a file that contains the MAHADISCOM Sub Engineer exam syllabus. You can download this from here. Some contains of the file is as follows:

Paper I - Total – 100 Marks Duration 3 Hours

A] The Indian Electricity Rules 1956 –[ On revision of IE Rules as per the provisions of
Electricity Act 2003 the same would be applicable].

B] The Electricity Act 2003 : ( as amended from time to time with following chapters.
Part-I : Preliminary.
Part-IV : Licensing.
Part-VI : Distribution of Electricity.
Part-VIII : Works.
Part-XII : Investigation and Penalties.
Part-XIV : Offences and Penalties.
Part-XV : Special Courts.
Part-XVI : Disputes. Resolutions and
Part-XVII : Other Provisions.

MAHADISCOM Sub Engineer exam syllabus are detail to atteched a pdf file.............

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MAHADISCOM Sub Engineer exam syllabus.pdf (114.8 KB, 70 views)

MAHADISCOM State Electricity Distribution Company Limited – MSEDCL is the second largest Electricity Distribution Company. The Company operates under Government of India.

Sample Questions:

1. Find the Fourier sine transform of f(x), where
f(x) = f(x) 1,0 { 0, x>a
(a) √2/p (cos st / s)
(b) √2/p (1 - cos as / s) (Ans)
(c) √2 (1 - cos as)
(d) None of these
2. A random variable X with uniform density in the interval 0 to 1 is Quantized as follows:
If 0 ≤ X ≤ 0.3, xq = 0
If 0.3 ≤ X ≤ 1, xq = 0.7
Where Xq is the quatized value of x
The root mean square value of the quantization noise is
(a) 0.573
(b) 0.198 (ans)
(c) 2.205
(d) 0.266
Solution : Since it is uniform as
xq = 0 in the range 0≤x≤0.3
xq = 0.7 in the range 0.3≤x≤1
The square mean value is
¥
s2 = ∫ (x - xq)2 f (x) dx
-¥
1
= ∫ (x - xq)2 f (x) dx
0
0.3 0.1
= ∫ (x - 0)2 f (x) dx + ∫ (x - 0.7)2 f (x) dx
0 0.3
0.3 1
= [x3/3] + [x3/3 + 0.49 x - 1.4]
0 0.3
or s2 = 0.039
The root mean square quantization noise
RMS = √s2
= √0.039 = 0.198


4. There analog signals, having bandwidth 1200 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with 12 bit words, and time division multiplexed. The bit rate for the multiplexed signal is
(a) 115.2 kbps
(b) 28.8 kbps
(c) 57.6 kbps (ans)
(d) 38.4 kbps
Solution : The three analog Signals having BW 1200 Hz, 600Hz and 600 Hz are sampled at their respective Nyquist rate i.e. at 2400, 1200, 1200 sample/sec respectively.
The total of (2400 + 1200 + 1200) = 4800 sample/sec
The Bit rate = n. fs = (4800 sample/sec) x 12 = 57.6 Kbps
Where n = number of bit in a symbol
6. Which of the following analog modulation scheme requires the minimum transmitted power and minimum channel bandwidth?
(a) VSB
(b) DSB-SC
(c) SSB (ans)
(d) AM
Solution : VSB → fm + fc
DBS - SC → 2 fm
SSB → fm
AM → 2 fm
Thus SSB has minimum bandwidth and it required minimum power i.e. 17% as compared to AM.
7. A device with input x(t) and output y(t) is characteristic by : y (t) = x2(t). An FM signal with frequency deviation of 90 KHz and modulation signal bandwidth of 5 KHz is applied to this device. The bandwidth of the output signal is
(a) 370 KHz (ans)
(b) 190 KHz
(c) 380 KHz
(d) 95 KHz
Solution : In present case
∆f = 90; fm = 5
β = [∆f / fm] = [90/5] = 18
FM equation
A cos [wct + β = sin wmt]
= A cos [wct + 18 sin wmt]
y(t) = x2 (t) = A2 cos2 [wct + 18 Sin wmt]
Note : Cos2 q = [1 + Cos2q] / 2
If there is change in frequency the modulation index also changes in same ratio
y(t) = A2 [(1/2) + (1/2) Cos {2wct + 36Sin wmt}]
y(t) = [(A2/2) + (A2/2) Cos {2wct + 36Sin wmt}]
After the device,
β(new) = 36 = [∆f(new) / fm]
∆f(new) = 36 x 5 = 180
By carson's rule
Bandwidth = 2(∆f + fm)
= 2 (180 + 5)
Bandwidth = 370 kHz
9. A carrier is phase modulated (PM) with frequency deviation of 10 KHz by a single tone frequency of 1 KHz. If the single tone frequency is increased to 2 KHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is
(a) 21 kHz
(b) 22 kHz
(c) 42 kHz
(d) 44 kHz (ans)
Solution : ∆f = 10 KHz fm(new) = 2 KHz
fm = 1 KHz
By carson's Rule
BW = 2 (∆f + fm) = 2 (10 + 1) = 22 KHz
∆ f(new) = 2 x 10 = 20
BW(new) = 2 (20 + 2) = 44 kHz
10 If A and B be the set and Ac and Bc denote the complements of the sets. A and B, then set (A - B) È (B - A) È (A Ç B) is equal to
(a) A È B (Ans)
(b) Ac È Bc
(c) A Ç B
(d) Ac Ç Bc

11 Let G = G(V, E) has five vertices, then the maximum number of m of edges in E, if G is a multigraph ?
(a) 5
(b) 2
(c) 10
(d) Finite or infinite (Ans)
12 How many straight line can be drawn through 10 points on a circle ?
(a) 10
(b) 20
(c) 45 (Ans)
(d) Infinite
13 . The Fourier transform of unit step function u(t) is
(a) 1
(b) pd(w)
(c) pd(w) - 1/jw (Ans)
(d) pd(w) + 1/jw

14. The value of the integral ∫ e-2(x - t) d(t - 2) dt is
-∞
(a) e-2(x - 2) (Ans)
(b) e2(x - 2)
(c) e-2(x + 2)
(d) e2(x + 2)

15. The uint of Ñ x H is
(a) A
(b) A/m
(c) A/m2 (Ans)
(d) A-m


Here I am attaching the sample papers of the MAHADISCOM Sub Engineer Exam:

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MAHADISCOM Sub Engineer Paper.doc (60.0 KB, 141 views)

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