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Q1. | In a synchro error detector, the output voltage is proportional to [ω(t)]^{n}, where ω(t) is the rotor velocity and n equals : |

A. | -2 [Wrong Answer] |

B. | -1 [Wrong Answer] |

C. | 1 [Correct Answer] |

D. | 2 [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 :**

⇒ Square law modulators utilise

linear range of V-I characteristics of a diode

linear range of V-I characteristics of a triode

non-linear region of V-I dynamic characteristics of a diode

non-linear region of V-I dynamic characteristics of a triode

⇒ A feedback amplifier has forward gains A

_{1}, A

_{2}and feedback factor β as shown in signal flow graph of the given figure. Because of feedback as shown in signal noise ratio at output

increase

decrease

be unaffected

depend on value of β

⇒ In a minimum phase system

all poles lie in the left half plane

all zeros lie in the left half plane

all poles lie in the right half plane

all except one pole or zero lie in the left half plane

⇒ In electromagnetic wave polarization is caused by

refraction

reflection

longitudinal nature of electromagnetic waves

transverse nature of electromagnetic waves

⇒ An

*npn*transistor has a Beta cutoff frequency

*f*

_{β}of 1 MHz, and a common emitter short circuit low frequency current gain β

_{0}of 200. It unity gain frequency

*f*

_{T}and the alpha cut off frequency

*f*

_{a2}respectively are

200 MHz, 201 MHz

200 MHz, 199 MHz

199 MHz, 200 MHz

201 MHz, 200 MHz

⇒ Transistor T

_{1}operates at 30 kHz and T

_{2}operates at 300 Hz. The flicker noise is

not depend on frequency

T

_{1}> T

_{2}

T

_{1}= T

_{2}

T

_{1}< T

_{2}

⇒ Radiowaves are electromagnetic waves having frequency range

0.001 to 50 H

0.001 to 50 kHz

0.002 to 50 MHz

0.001 to 10

^{16}Hz

⇒ The intermediate frequency of a superheterodyne receiver is 450 kHz. If it is tuned to 1200 kHz, the image frequency will be

750 kHz

900 kHz

1600 kHz

2100 kHz

⇒ When a

*p*-

*n*junction is reverse biased

holes and electrons move away from the junction

holes and electrons move towards the junction

holes move towards junction and electrons move away from junction

holes move away from junction and electrons move towards junction

⇒ A Darlington pair has β = 300. If R

_{E}= 1 kΩ and R

_{L}= 100 Ω, the input impedance at the base is approximately

300 k ohm

3 k ohm

100 k ohm

30 k ohm

⇒ The effective value of the waveform in the given figur

5

2.5

2.5

5.0

⇒ In the circuit shown in the figure, the current supplied by the sinusoidal current source

28 A

0.166666666666667

20 A

not determinable from the data given

⇒

Match the following:

List I | List II | ||
---|---|---|---|

A. | Resistance | 1. | M^{-1}L^{-2}T^{2}I^{2} |

B. | Inductance | 2. | ML^{2}T^{-2}I^{-2} |

C. | Capacitance | 3. | M^{-1}L^{-2}T^{4}I^{2} |

D. | Reluctance | 4. | ML^{2}T^{-3}I^{-2} |

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

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

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

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

⇒ The

*z*-transform of sequence

*x*[

*n*] = δ(

*n*) is

1

∞

0

*z*

⇒ The number of branches of root locus plot is equal to

the number of roots of characteristic equation

double the number of roots of characteristic equation

the number of roots of characteristic equation minus one

the number of roots of characteristic equation plus one

⇒ In the circuit shown in given figure, the values of i(0

^{+}) and I(∞), will be respecti

zero and 1.5 A

1.5 A and 3 A

3 A and zero

3 A and 1.5 A

⇒ The maximum effective aperture of a microwave antenna which has a directivity of 900, will be

71.619 λ

^{2}

716.19 λ

^{2}

7161.9 λ

^{2}

71619 λ

^{2}

⇒ FM receivers using the standard 88 to 108 MHz band use IF of

8 MHz

9.9 MHz

10.7 MHz

12.2 MHz

⇒ The quality of a space-link is measured in terms of the __________ ratio.

C/N

S/N

G/T

EIRP

⇒ The trapezoidal pattern of figure is obtained when examining an AM wave by an oscilloscope. Modulation factor of the wave

0.33

0.5

2

1

⇒ The temperature of cathode is increased from 2500K to 2600K. The increase in thermionic emission current is about

0.001

0.04

0.5

1.5

⇒ Wire-wound resistors are unsuitable for use at high frequencies because they

are likely to melt under excessive eddy current heat

exhibit unwanted inductive and capacitive effects

create more electrical noise

consume more power

⇒ A parabolic antenna produces a radiation pattern which is

circularly polarized

highly directional

omnidirectional

linearly polarized

⇒ Consider the following statements

- Usable bandwidth of ridge wave guide is more than that of rectangular waveguide.
- For a given value of TM
_{10}cutoff frequency the cross section of ridge wave guide is lesser that that of rect angular waveguide. - Ridge wave guide has higher conductor losser than a rectangular waveguide.
- Ridge guide has higher conductor losser than a rectangular waveguide.

1 and 2 only

1, 2 and 3 only

2 and 3 only

All

⇒ Master control facility (MCF) for INSAT-2 series satellites is located at

Madras

New Delhi

Leh

Hassan

⇒ For a message signal

*m*(

*t*) = cos (2p

*f*) and the carrier frequency

_{m}t*f*, which of the following represents a single side band (SSB) signal?

_{c}cos (2p

*f*) cos (2p

_{m}t*f*)

_{c}tcos (2p

*f*)

_{c}tcos [2p(

*f*+

_{c}*f*) t]

_{m}[1 + cos (2p

*f*) cos (2p

_{m}t*f*)]

_{c}t⇒ In a driving point function, the degrees of numerator and denominator polynomials in

*s*may differ by either zero or one only.

TRUE

FALSE

⇒ To draw dc equivalent circuit for a CE amplifier circuit we should

Reduce all ac sources to zero

Open all capacitors

Reduce all ac sources to zero and open all capacitors

Reduce all ac sources to zero and short all capacitors

⇒ For aeroplane and navigation frequency band is

VLF

UHF

VHF

EHF

⇒ The circuit shown in the figure conv

BCD to Binary code

Binary to excess - 3 code

Excess - 3 to Gray

Gray to Binary