VIT Engineering VIT Engineering Solved Paper-2007

  • question_answer
    The transmission of high frequencies in a coaxial cable is determined by:

    A)  \[\frac{1}{{{\left( LC \right)}^{1/2}}},\], where L and C are inductance and capacitance

    B)  \[{{\left( LC \right)}^{2}}\]

    C)  the impedance L alone

    D)  the dielectric and skin effect

    Correct Answer: D

    Solution :

    Basically, a coaxial cable consists of a hollow (outer) cylindrical conductor surrounding a single (inner) conductor along its axis. The two conductors are well insulated from each other. The electric field \[(\overrightarrow{E})\] and magnetic field \[(\overrightarrow{H})\] at the cross-sections are shown by solid lines and dotted lines, respectively. The   outer conductor acts as the shield and minimises interference. Different kinds of dielectric materials, such as teflon and polythene are covered over copper wire, it acts as a spacer. In the transmission of power through coaxial cable, the dielectric medium separating the inner conductor from outer one plays a vital role. These dielectric materials are good insulators only at low frequencies. As the frequency increases, the energy loss becomes significant. That is why a coaxial cable can be used effectively for transmission upto a frequency of 20 MHz. A steady signal flowing in a wire, uniformly distributes itself throughout the cross-section of the wire. A high frequency signal, on the other hand distributes itself uniformly, there being a concentration of current on the outer surface of the conductor. If the frequency of the current is very high, the current is almost wholly confined to the surface layers. This is called Skin effect.


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