Скачать или смотреть TRB Polytechnic | 2017-18 | Unit 9 | Electronics | கேட்கப்பட்ட கேள்விகள் | SS Academy for physics

TRB Polytechnic | 2017 -18 | Unit 9 | Electronics | MCQ | கேட்கப்பட்ட கேள்விகள் | Detailed Explanation | SS Academy for physics
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Multiple choice questions in trb physics
mcq trb polytechnic
Multiple choice questions in trb polytechnic
Electronics mcq
Electronics Multiple choice questions
1). A logic gate whose output is in '1' state if and only if all the inputs are in '1' state is :
AND gate
NAND gate
NOT gate
OR gate
2) An ideal operational amplifier has :
finite input impedance
finite bandwidth
infinite input impedance
zero input impedance
3). In an Op-amp the input terminal marked as '-' (minus sign) is called as :
(A) inverting input terminal
(B) phase change terminal
(C) inverting output terminal
(D) non-inverting terminal
4) . Increase in current in tunnel diode from 0 to its peak value IP at a forward voltage VP, is due to the phenomena called :
(A) tunneling (B) breakdown
(C) flow of electrons in reverse (D) negative resistance
5). Convert (25)10 into binary numbers :
(A) (0110)2
(B) (1011)2
(C) (11101)2
(D) (11001)2
6). The expression for De Morgan's theorem is :
(A) AB = 𝐴 ̅ + 𝐵 ̅
(B) (𝐴𝐵) ̅ = 𝐴 ̅ + B
(C) (𝐴𝐵) ̅ = 𝐴 ̅ + 𝐵 ̅
(D) (𝐴𝐵) ̅ = 𝐴 ̅ . 𝐵 ̅
7). In a phase shift oscillator, a ladder network of three RC section is used each section will produce a phase shift of :
90o
40o
60o
70o
8). If C1R2 is the time constant, the width of the output pulse T in monostable vibrator :
T = 0.89 C1R2
T = 0.59 C1R2
T = 0.79 C1R2
T = 0.69 C1R2
9) For a sustained oscillations in phase shift oscillator, the minimum value of the forward current gain ratio of the transistor should be :
46
66
38
56
10). If the tunnel diode is reverse biased it acts like a:
Conductor
Superconductor
Semiconductor
Insulator
12). A 5-variable K-map has :
64 cells
16 cells
32 cells
5 cells
13). Simplify Using K-Map Y = F(A,B,C,D) = ∑▒〖 3, 4, 5, 7, 9, 13, 14, 15〗

Y = 𝐴 ̅ B 𝐶 ̅ + 𝐴 ̅ CD + A 𝐶 ̅ D + ABC + BD
Y = AB 𝐶 ̅ + A 𝐶 ̅ D + 𝐴 ̅ 𝐶 ̅ D + 𝐴 ̅( 𝐵) ̅( 𝐶) ̅
Y = (𝐴𝐵) ̅ C+ 𝐴 ̅ C 𝐷 ̅ + 𝐴 ̅ 𝐶 ̅ D + 𝐴 ̅ BC
Y = 𝐴 ̅ B 𝐶 ̅ + 𝐴 ̅ CD + A 𝐶 ̅ D + ABC