MPPKVVCL Electrical-Engineering-EE Contributed by Swathy updated on May 2019

http://p3.placement.freshersworld.com/placement-papers/
MPPKVVCL Selection process
1.Written test
2.Interview

a)  The Eligible candidates shall have to undergo a written test comprising of questions in relevant branch of Engineering and General Aptitude (Logical Reasoning and General Knowledge).

b)  The written test will be of 2 hours 30 minutes duration.

c)   Date of written test 24th March 2013

MPPKVVCL electronics and electrical engineering model question for practice.This is only sample placement papers

1.  To neglect a voltage source, the terminals across the source are

(A)  open circuited

(B)  short circuited     (Ans)

(C)  replaced by some resistance

(D)  replaced by an inductor

2. Current I0 in the given circuit will be

(A)  10 A

(B)  3.33 A     (Ans)

(C)  20 A

(D)  2.5 A

Solution :  RT = 2 + 2 || [1 + (2 || 2)]

= 2 + 2 || (1 + 1)

= 2 + 2 || 2 = 2 + 1 = 3 ?

So    I =  40 A
3

By current division I0 = 1 * 1 * 40  =  10 A
2    2    3        3

I0 = 3.33 A

3.  In a resonant circuit, the power factor at resonance is

(A)  zero

(B)  unity     (Ans)

(C)  0.5

(D)  1.5

4.  In the given circuit voltage V is reduced to half. The current will become

(A)  I/2    (Ans)

(B)  2 I

(C)  1.5 I

(D)  I / ?R2 + (XL + XC)2

Solution :   I a V if V reduced half current becomes   I .
2

5.  The function          3 s           has
(s + 1) (s + 2)

(A)  one zero, two poles    (Ans)

(B)  no zero, one pole

(C)  no zero, two poles

(D)  one zero, no pole

Solution :  G (s) =           3s            has one zero at s = 0 and two poles at s =  - 1,  - 2.
(s + 1) (s + 2)

6.  One electron volt is equivalent to

(A)  1.6 * 10-10 J

(B)  1.6 * 10-13 J

(C)  1.6 * 10-16 J

(D)  1.6 * 10-19 J    (Ans)

7.  Which of the following is donor impurity element ?

(A)  Aluminium

(B)  Boron

(C)  Phosphorous    (Ans)

(D)  Indium.

8.  The diameter of an atom is of the order

(A)  10-6 m

(B)  10-10 m   (Ans)

(C)  10-15 m

(D)  10-21 m

9.  The following Figure represents

Image - BSNL- MB (J.T.O.) RE 2001 - Part A - Qno. 9

(A)  LED    (Ans)

(B)  Varistor

(C)  SCR

(D)  Diac

10.  If the d.c. valve of a rectified output is 300 V and peak to peak ripple voltage is 10 V, the ripple factor is

(A)  1.18%

(B)  3.33%   (Ans)

(C)  3.36%

(D)  6.66%

Solution :  rms value of output

= ?3002 + 102 = 300.166 V

Average value = 300 V

Form factor      RMS value     = 300.166 = 1.00055.
Average value          300

Ripple factor = ?(Form factor)2 - 1

= ?(1.0005)2 - 1 = 0.0333

= 3.33%.

11.  Full wave rectifier output has ripple factor of

(A)  1.11    (Ans)

(B)  1.21

(C)  1.41

(D)  1.51

12.  In a common base connection IE = 2 mA, IC = 1.9 mA. The value of base current is

(A)  0.1 m    (Ans)

(B)  0.2 mA

(C)  0.3 mA

(D)  zero

Solution :  IE = 2 mA     IC = 1.9 mA

Ib = IE  - IC = (2 - 1.9) = 0.1 mA.

13.  For the action of transistor the base region must be

(A)  P-type material

(B)  N-type material

(C)  very narrow     (Ans)

(D)  highly doped

14.  Compared to a CB amplifier the CE amplifier has higher

(A)  current amplification

(B)  output dynamic resistance

(C)  leakage current

(D)  input dynamic resistance

(E)  all of the above    (Ans)

15.  When a transistor is biased to cut-off its Y is

(A)  0.5

(B)  0

(C)  1.0    (Ans)

(D)  0.8

16.  An ideal voltage amplifier should have

(A)  Ri = 0,  R0 = 0

(B)  Ri = 0,  R0 = ¥

(C)  Ri = ¥,  R0 = 0    (Ans)

(D)  Ri = 0,  R0 = ¥

17.  Barichausen criterion for sustained oscillator is

(A)  A? = 1 Ð00    (Ans)

(B)  A? = 0

(C)  A? = 1 Ð1800

(D)  A = 1/??

18.  The value of plastic capacitance for a triode may range from

(A)  2 ?F to 12 ?F

(B)  20 ?F to 120 ?F

(C)  2 pF to 12 pF    (Ans)

(D)  20 pF to 120 pF

19.  A colpitts oscillator uses

(A)  tapped coil

(B)  inductive feedback

(C)  tapped capacitance     (Ans)

(D)  no tuned LC circuit

20.  A typical frequency for a RC feedback oscillator is

(A)  1 KHz     (Ans)

(B)  100 MHz

(C)  1000 MHz

(D)  1 GHz