Users Also Read

MCQ's Search Engine

Electrical Engineering

Mechanical Engineering

Civil Engineering

Automobile Engineering

Chemical Engineering

Computer Engineering

Electronics Engineering

Medical Science Engg

Q1. | A full wave bridge rectifier is supplied voltage at 50 Hz. The lowest ripple frequency will be |

A. | 400 Hz [Wrong Answer] |

B. | 200 Hz [Wrong Answer] |

C. | 100 Hz [Correct Answer] |

D. | 50 Hz [Wrong Answer] |

View Answer
Explanation:-
Answer : CDiscuss 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 :**

⇒ Consider the following; for a loss-less transmission line we can write,

- Z
_{in}= -*jz*_{0}for a shorted line with*l*= λ/8 - Z
_{in}= ±*j*∞ for a shorted line with*l*= λ/4 - Z
_{in}=*jz*_{0}for an open line with*l*= λ/2 - Z
_{in}= 0 for a matched line of any length

1, 2

2, 3

1, 3

2, 4

⇒ Rank of the matr

3

2

1

4

⇒ In 8085, F register has 8 bits.

TRUE

FALSE

⇒ A pre-emphasis circuit provides extra noise immunity by

converting the phase modulation to frequency modulation

pre-amplifying the whole audio band

boosting the base frequencies

amplifying the higher audio frequencies

⇒ Find the FSV (full scale voltage) in a 6 bit R-2

*R*ladder D/A converter has a reference voltage of 6.5 V.

6.4 V

0.1 V

7 V

8 V

⇒ There are four different algorithms A1, A2, A3 and A4 to solve, a given problem with the complexity order log(n), log(log(n)), n log and n/log(n) respectively. Which is the best algorithm?

A1

A2

A3

A4

⇒ A potential of 7 V is applied to a silicon diode. A resistance of 1 K ohm is also in series with the diode. The current is

7 mA

6.3 mA

0.7 mA

0

⇒ In a

*p*type material the Fermi level is 0.3 eV above valence band. The concentration of acceptor atoms is increased. The new position of Fermi level is likely to be

0.5 eV above valence band

0.28 eV above valence band

0.1 eV above valence band

below the valence band

⇒ In 8085, the instruction CMA is an example of

Implict addressing

Direct addressing

Register addressing

Immediate addressing

⇒ For the transport lag G(

*j*ω) =

*e*

^{-jωT}, the magnitude is always equal to

0

1

10

0.5

⇒ In figure

*v*

_{1}= 8 V and

*v*

_{2}= 4 V. Which diode will cond

D

_{2}only

D

_{1}only

Both D

_{1}and D

_{2}

Neither D

_{1}nor D

_{2}

⇒ In a pulsed RADAR typical value of pulse duration and pulse frequency respectively are

1 μs and 50 Hz

1 μs and 500 Hz

4 μs and 50 Hz

4 μs and 500 Hz

⇒ -8 is equal to signed binary number

10001000

1000

10000000

11000000

⇒ In figure the voltage drop across the 10 Ω resistance is 10 V. The resistan

is 6 Ω

is 8 Ω

is 4 Ω

cannot be found

⇒ If luminance signal Y = 0.3 R + 0.59 G + 0.11 B, the hue of R - Y colour is

0.7 R - 0.59 G - 0.11 B

- 0.7 R + 0.59 G + 0.11 B

1.3 R - 0.59 G - 0.11 B

none of the above

⇒ The superposition theorem is applicable to

linear, non-linear and time variant responses

linear and non-linear resistors only

linear responses only

none of the above

⇒ In a superheterodyne receiver, the IF is 455 kHz. If it is tuned to 1200 kHz, the image frequency will be

1655 kHz

2110 kHz

745 kHz

910 kHz

⇒ A (0.75 -

*j*50) ohm load is connected to a coaxial transmission line of Z

_{0}= 75 ohms at a frequency of 10 GHz Which of the following will be the best method of matching?

Connecting an inductance at the end

Connecting a short circuited stub at the load

Connecting a capacitance at some distance from the load

Connecting a short circuited stub at some specified distance from the load

⇒ In a single phase cycloconverter with centre tapped transformer

all thyristors carry currents for equal duration

the durations of currents in all thyristors may or may not be equal

the durations of currents in all thyristors cannot be equal

the duration of currents in all thyristors must be equal for satisfactory operation

⇒ Four messages band limited to W, W, 2W and 3W respectively are to be multiplex using Time Division Multiplexing (TDM). The minimum bandwidth required for transmission of this TDM signal is

W

3W

6W

7W

⇒ Consider the following program in Ba

60, 20, 20, 30, 10

10, 30, 20, 20, 60

60, 20, 30, 10, 20

60, 30, 10, 20, 20

⇒ A signal

*x*

_{1}(

*t*) and

*x*

_{2}(

*t*) constitute the real and imaginary parts respectively of a complex valued signal

*x*(

*t*). What form of waveform does

*x*(

*t*) possess?

Real symmetric

Complex symmetric

Asymmetric

Conjugate symmetric

⇒ Tellegen's theorem is applicable to

circuits with passive elements only

circuits with linear elements only

circuits with time invariant elements only

circuits with active or passive, linear or nonlinear and time invariant or time varying elements

⇒

Match the following:

List I (Logic family) | List II (Characteristic of logic family) | ||
---|---|---|---|

A. | TTL | 1. | Maximum power Consumption |

B. | ECL | 2. | Highest Packing Density |

C. | NMOS | 3. | Least Power Consumption |

D. | CMOS | 4. | Saturated Logic |

A-1, B-4, C-2, D-3

A-1, B-4, C-3, D-2

A-4, B-1, C-2, D-3

A-4, B-1, C-3, D-2

⇒ Insulation resistance of a cable is directly proportional to its length.

TRUE

FALSE

⇒ Which of the following is not a valid real constant in C?

426

321

-621.231

201.2314

⇒

Match the following:

List I (Application of circuit) | List II (Circuit Name) | ||
---|---|---|---|

A. | Divider | 1. | Astable multivibrator |

B. | Clips input voltage at two predetermined levels | 2. | Schmitt trigger |

C. | Square wave generator | 3. | Bistable multivibrator |

D. | Narrow current pulse generator |

A-4, B-2, C-1, D-3

A-3, B-2, C-1, D-4

A-4, B-1, C-2, D-3

A-3, B-1, C-2, D-4

⇒ Consider the following statement in FORTRAN 77

4 values could be read and stored asX (0), Y (0), X (2), Y (2)

6 values could be read and stored as X (0), Y (0), Y (2), X (3), Y (3)

none of the above

⇒ The circuit in figur

Cauer I form

Foster I form

Cauer II form

Foster II form

⇒ The fourier transform of the half cosine pulse as shown below is _______

0.5 {sin [0.5(

*f*- 1)] + sin [0.5 (

*f*+ 1)]}

0.25 [sin c[0.25 (

*f*- 2)1 + sin c[0.25(

*f*+ 2)]]

0.5 {sin c [0.5 (

*f*- 1)] + sin c [0.5 (

*f*+ 1)]}

sin c(

*f*- 0.5) + sin c (

*f*+ 0.5)