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Q1. | In a coaxial cable, braided copper is used for |

A. | conductor [Wrong Answer] |

B. | shield [Correct Answer] |

C. | dielectric [Wrong Answer] |

D. | jacket [Wrong Answer] |

View Answer
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 :**

⇒ In the circuit shown below, the outputs Y

_{1}and Y

_{2}for the given initial condition Y

_{1}= Y

_{2}= 1 and after four input pulses wil

Y

_{1}= 1, Y

_{2}= 0

Y

_{1}= 0, Y

_{2}= 0

Y

_{1}= 0, Y

_{2}= 1

Y

_{1}= 1, Y

_{2}= 1

⇒ In figure, the current through 10 V batter

0

0.5 A charging

0.5 A discharging

none of the above

⇒ It is desired to couple a coaxial line to a two parallel wire line. It is best to use

quarter wave transformer

balun

slotted line

directional coupler

⇒ In cellular telephone each cell is designed to handle

45 two way communications

90 two way communications

45 one way communications

180 two way communications

⇒ For an I/p of V

_{s}= 5 sin ω

*t*(assuming ideal diode), circuit shown in the figure will behave

clipper, sine wave clipped at -2 V

clamper, sine wave clamped at -2 V

clamper, sine wave clamped at 0 volts

clipper, sine wave clipped at 2 V

⇒ Some sources of TV programs are

- TV camera
- Telecine
- VTR
- External signal

1, 2, 3 and 4

1, 2, 3

1 and 3

1 2 and 4

⇒ The presence of some holes in an intrinsic semiconductor at room temperature is due to

valence electrons

doping

free electrons

thermal energy

⇒ A speech signal, band limited to 4kHz and peak voltage varying between + 5V and - 5V is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. The number of quantization levels required to reduce the quantization noise by a factor of 4 would be

1024

512

256

64

⇒ In a linear circuit, the superposition principle can be applied to calculate the

voltage and power

voltage and current

current and power

voltage, current and power

⇒ A parallel RLC circuit has ω

_{0}= 10

^{6}and Q = 30, Given C = 30 pF, the value of R is

0

1 μΩ

2 MΩ

1 MΩ

⇒ For a superconductor, magnetisation M in a field H is

extremely small

- H

- I

zero

⇒ The collector voltage V

_{C}of the circuit shown in the given figure aside is approxima

2 V

4.6 V

8 V

8.6 V

⇒ The distance between target and radar is reduced to half. Power received increases by

42751

42739

4 times

16 times

⇒ The mean free path of conduction electrons in copper is about 4 x 10

^{-8}m. For a copper block, find the electric field which can give, on an average, 1 eV energy to a conduction electron

2.62 x 10

^{7}V/m

2.64 x 10

^{7}V/m

2.5 x 10

^{7}V/m

2.58 x 10

^{7}V/m

⇒ The emf of a Weston cell is 1.286 V.

TRUE

FALSE

⇒ A ripple counter has propagation delay.

TRUE

FALSE

⇒

**Assertion (A):** For a physically realisable driving point function, the degree of numerator and denominator should be equal.

**Reason (R):** The sum of positive real functions is real.

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

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

A is true but R is false

A is false but R is true

⇒ Poynting vector signifies which of the following?

Power density vector producing electrostatic field

Power density vector producing electromagnetic field

Current density vector producing electrostatic field

Current density vector producing electromagnetic field

⇒ ADM stands for

Angle Delta Modulation

Angle Digital Modulation

Adaptive Delta Modulation

Adaptive Digital Modulation

⇒ If the absolute value of a number is less than decimal 1, the sign of exponent in floating point representation is

zero

negative

zero or negative

positive

⇒ The unit of loudness is

phon

sone

decibel

mel

⇒ A circular waveguide carries TE

_{11}mode whole radial electric field is give

10 cm

3 p cm

2 p cm

8 cm

⇒ The Fourier series of an odd periodic function contains

odd harmonics only

even harmonics only

cosine harmonics only

sine harmonics only

⇒ In a power diode the reverse recovery time is the time from the instant the forward current is zero to the instant when reverse recovery current has decayed to

50% of peak reverse current

25% of peak reverse current

10% of peak reverse current

zero value

⇒ An SCR is triggered at 80° in the positive half cycle, only. The rms anode current is 25 A. If the firing angle is changed to 160°, the rms current is likely to be

25 A

12.5 A

less than 25 A

less than 12.5 A

⇒ A 1000 kHz carrier is simultaneously modulated with 300 Hz, 800 Hz and 2 kHz audio sine waves. The frequencies present in the output will be

998 kHz and 1002 kHz

998 kHz, 999.2 kHz, 1000.8 kHz, 1002.0 kHz

998 kHz, 999.2 kHz, 999.7 kHz, 1000.3 kHz, 1000.8 kHz and 1000.2 kHz

none of the above

⇒ Piezoelectric quartz crystal resonators find application where

signal amplification is required

rectification of the signal is required

signal frequency control is required

modulation of signal is required

⇒ In a BJT circuit a

*pnp*transistor is replaced by

*npn*transistor. To analyse the new circuit

all calculations done earlier have to be repeated

replace all calculated voltages by reverse values

replace all calculated currents by reverse values

replace all calculated voltages and currents by reverse values

⇒ The attenuation factor in parallel plane guides is given by

a = power lost/power transmitted

a = 2 x power lost/power transmitted

a = power lost per unit length/2 x power transmitted

a is always constant

⇒ The circuit of the figure is an example of feedback of the following

current series

current shunt

voltage series

voltage shunt