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)
(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
(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