Electromagnetic theory miscellaneous


Electromagnetic theory miscellaneous

Electromagnetic Theory

  1. If the diameter of a λ/2 dipole antenna is increased from λ/100 to λ/50 then its—









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    See from synopsis provided that smaller diameter result in narrower bandwidth and larger diameter gives wider bandwidth.

    Correct Option: A

    See from synopsis provided that smaller diameter result in narrower bandwidth and larger diameter gives wider bandwidth.


  1. The frequency range for satellite communication is—









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    NA

    Correct Option: D

    NA



  1. A-TEM wave incident normally upon a perfect conductor the E and H fields at the boundary will be respectively—









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    At boundary the fields are in phase opposition 180º out of phase and whole energy is reflected back for perfect conductor. Hence E and H fields at the boundary will be maximum and minimum respectively.

    Correct Option: D

    At boundary the fields are in phase opposition 180º out of phase and whole energy is reflected back for perfect conductor. Hence E and H fields at the boundary will be maximum and minimum respectively.


  1. The magnitude of open circuit and short circuit input impedances of a transmission line are 100 Ω and 25 Ω respectively the characteristic impedance of the line is—









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    Characteristic impedance is given in terms of short circuit and open circuit impedance Zc =√ZSCZOC = √25 × 100 = √2500 = 50Ω

    Correct Option: B

    Characteristic impedance is given in terms of short circuit and open circuit impedance Zc =√ZSCZOC = √25 × 100 = √2500 = 50Ω



  1. Find Q of the coil at resonance having frequency 1 MHz—









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    At resonance XL = XC

    jωL =
    1
    jωC

    or
    L =
    1
    =
    1
    ω20C(2∏ × 106)2 × 50 × 10−12

    =
    1
    50 × 4∏2
    Q =
    ω20L
    =
    106 × 2∏
    ×
    1
    R1050 × 4∏2

    =
    103

    Correct Option: A

    At resonance XL = XC

    jωL =
    1
    jωC

    or
    L =
    1
    =
    1
    ω20C(2∏ × 106)2 × 50 × 10−12

    =
    1
    50 × 4∏2
    Q =
    ω20L
    =
    106 × 2∏
    ×
    1
    R1050 × 4∏2

    =
    103