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SAIL Electronics and Electrical engineering objective type questions with answers for practice, SAIL model questions for practice,SAIL free solved sample placement papers of previous years 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 I 0 in the given circuit will be (A) 10 A (B) 3.33 A (Ans) (C) 20 A (D) 2.5 A Solution : R T = 2 + 2 || [1 + (2 || 2)] = 2 + 2 || (1 + 1) = 2 + 2 || 2 = 2 + 1 = 3 Ω So I = 40 A 3 By current division I 0 =1 *1 * 40 = 10 A 2 2 3 3 I 0 = 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

SAIL Electronics and Electrical Engineering Objective Type Questions With Answers for Practice

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Page 1: SAIL Electronics and Electrical Engineering Objective Type Questions With Answers for Practice

SAIL Electronics and Electrical engineering objective type questions with answers for practice, SAIL model questions for practice,SAIL free solved sample placement papers of previous years

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 Ω

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)

Page 2: SAIL Electronics and Electrical Engineering Objective Type Questions With Answers for Practice

(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)   =                            3   s                                 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

(A)  LED    (Ans)

Page 3: SAIL Electronics and Electrical Engineering Objective Type Questions With Answers for Practice

(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)

Page 4: SAIL Electronics and Electrical Engineering Objective Type Questions With Answers for Practice

(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