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Q1. | For a bipolar transistor, avalanche multiplication factor depends on __________ . |

A. | V_{CE} [Wrong Answer] |

B. | V_{CB} [Correct Answer] |

C. | V_{BE} [Wrong Answer] |

D. | none of these [Wrong Answer] |

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Explanation:-
Answer : BDiscuss it below :!! OOPS Login [Click here] is required to post your answer/resultHelp other students, write article, leave your comments |

**Also Read Similar Questions Below :**

⇒ Program counter in a digital computer

counts the number of programs run in the machine

counts the number of times a subroutine

counts the number of time the loops are executed

points the memory address of the current or the next instruction

⇒ The transfer function G

_{12}(

*s*) for a two port is

V

_{2}/V

_{1}

V

_{2}(

*s*)/V

_{1}(

*s*)

V

_{2}(

*s*)/I

_{1}(

*s*)

I

_{2}(

*s*)/I

_{1}(

*s*)

⇒ In a 4 input OR gate, the total number of High outputs for the 16 input states are

16

15

14

13

⇒ The generally used value of EHV supply in TV receivers is

1 kV per diagonal cm

1 kV per diagonal inch

0.5 kV per diagonal cm

0, 5 kV per diagonal inch

⇒ Consider an input

*x*(

*t*) as sh

-2 <

*t*< 6

0 <

*t*< 11

-2 <

*t*< 5

2 <

*t*< 6

⇒ Gold is often diffused into silicon PN junction devices to

increase the recombination rate

reduce the recombination rate

make silicon a direct gap semiconductor

make silicon semimetal

⇒ For a rectangular waveguide (

*a*x

*b*,

*a*>

*b*) to support only the TE

_{10}mode at wavelength which one of the following pairs of in equalities is to be satisfied?

*b*< λ < 2

*b*; λ > 2

*a*

*b*< λ < 2

*b*; λ < 2

*a*

*a*< λ < 2

*b*; λ < 2

*a*

*a*< λ < 2

*b*; λ > 2

*a*

⇒ Consider the network: Impedance of this network as a function of the complex frequency is consists of a certain number of zeros and poles. What is the location of po

-2

-2, ∞

2

2, ∞

⇒ The open loop gain of an amplifier is 50 but likely to decrease by 20% due to various factors. If negative feedback with β = 0.1 is used, the change in gain will be about

0.2

0.02

0.005

0.0005

⇒ In

*p*channel JFET, V

_{GS}is positive.

TRUE

FALSE

⇒ The carrier frequency will be if L = 50 µH and C = 1 nF.

512

612 kHz

712 kHz

812 kHz

⇒ Binary 1000 multiplied by binary 1000 gives

10000

100000

1000000

10000000

⇒ Which of the following pairs of octal and binary numbers are not equal?

111 11011

_{2}= 767

_{8}

110110101

_{2}= 665

_{8}

11010

_{2}= 3

_{8}

10101.11

_{2}= 25.6

_{8}

⇒ The circuit shown I/P in the figure repres

Low pass filter

High pass filter

Band Pass filter

Band stop filter

⇒ Polar plot of a lag-lead compensator is a circle.

TRUE

FALSE

⇒

**Assertion (A):** For elemental dielectrics, P = Na_{e}E_{i} where *N* is number of atoms/m^{3}, P is polarization, a_{e} is electronic polarization and E_{i} is internal electric field.

**Reason (R):** In elemental dielectrics there are no permanent dipoles or ions.

Both A and R are true and R is correct explanation of A

Both A and R are true but R is not correct explanation of A

A is true but R is false

A is false but R is true

⇒ A 12 bit ADC is used to convert analog voltage of 0 to 10 V into digital. The resolution is

2.44 mV

24.4 mV

1.2 V

none of the above

⇒ The radix of a hexadecimal system is

2

3

8

16

⇒ To prevent a DC return between source and load, it is necessary to use

resistor between source and load

inductor between source and load

capacitor between source and load

either (a) or (b)

⇒ When .4546 E 5 and .5433 E 7 are to be added in normalized floating point mode

none of the numbers is changed to any other form

.4546 E 5 is changed .004546 E 7 and .5433 E 7 is not changed

.5433 E 7 is changed to 54.33 E 5 and .4546 E 5 is not changed

both the numbers are changed and their exponents are made equal to 6

⇒ Consider the following program in Pascal

2.6

2.63

26.3

⇒ The 2 out of 5 code has

two 0s and three 0s in each code group

one 1s and four 0s in each code group

four 1s and one 0s in each code group

three 1s and two 0s in each code group

⇒ The transfer function of transportation lag is

G(

*j*ω) =

*e*

^{jωT }

G(

*j*ω) =

*e*

^{-jωT }

G(

*j*ω) = (1 +

*j*ωT)

G(

*j*ω) = (1 -

*j*ωT)

⇒ If V = 20 + j20, then log

_{e}(A) A is

3.342 +

*j*45

28.28 ∠45°

400 ∠45°

3.342 +

*j*0.785

⇒ Consider the following stateme

1, 4 are correct

1 and 2 correct

1, 2, 3 are correct

2, 3, 4 are correct

⇒ In the given figure the input frequency is such that

*R*<<

*X*,

_{c}*v*

_{0}lags

*v*

_{i}by 90°

*v*

_{0}leads

*v*

_{i}by 90°

*v*

_{0}and

*v*

*are in phase*

_{i}either (b) or (c)

⇒ Which pair is correctly matched?

Symmetrical two port AD - BC = 1

Reciprocal two port

*z*

_{11}=

*z*

_{12}

Inverse hybrid parameter A, B, C, D

hybrid parameter (V

_{1}, I

_{2}) =

*f*(I

_{1}, V

_{2})

⇒ In a mod-12 counter the input clock frequency is 10 kHz. The output frequency is

0.833 kHz

1.0 kHz

0.91 kHz

0.77 kHz

⇒ Consider the following statements

- Acceptor level lies close the valence band.
- Donor level lies close to the valence band.
*n*type semiconductor behaves as an insulator at 0 K.*p*type semiconductor behaves as an insulator at 0 K.

2 and 3 are correct

1 and 3 are correct

1 and 4 are correct

3 and 4 are correct

⇒ Which of the following compensation method in amplifier can be used to reduce the bandwidth?

Pole-zero compensation

Dominant pole compensation

Lead compensation

Any of the above