Electromagnetic theory miscellaneous
- A material is described by following electrical parameters as a frequency of 10 GHz, σ = 106 mho/m µ = µ0 and ε/ε0 = 10. The material at this frequencies is considered to be—
σ0 = 1 ×10−9 F/m 36 π
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Material may be found either conductor or insulator by the method of loss tangent
σ = 106 ωε 10 × 109 × 10 × 8.85 × 10-12
= 1.12 × 106
so σ >> ωεCorrect Option: A
Material may be found either conductor or insulator by the method of loss tangent
σ = 106 ωε 10 × 109 × 10 × 8.85 × 10-12
= 1.12 × 106
so σ >> ωε
- In a multicavity magnetron strapping is employed primarily—
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To prevent mode jumping and increase separation between the resonant frequencies in the π mode and in the adjacent modes.
Correct Option: E
To prevent mode jumping and increase separation between the resonant frequencies in the π mode and in the adjacent modes.
- The beamwidth between first null of uniform linear array of N equally spaced (element spacing = a) equally excited antennas is determined by—
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Beamwidth between first nulls is given by relation as follows
φ = sin-1 λ nd
where n = no. of sources, separated by d distance apart.Correct Option: D
Beamwidth between first nulls is given by relation as follows
φ = sin-1 λ nd
where n = no. of sources, separated by d distance apart.
- An open wire transmission line has primary constant
R = 20 Ω/m G = 0.3 × 103 s/km
L = 0.25 1 m H/km C = 0.1 µ F/m
at a frequency of 1 kHz its characteristic impedance is—
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We have characteristic impedance as follows
Z0 = √ R + jωL G + jωc
R = 20 Ω/ km,
G = 0.3 × 103 s/ km
L = 0.25 mH/ km,
C = 0.1 µF and ω = 103 Hz so we getZ0 = √ 20 + j2π(103 × 0.25 × 10−3) 0.3 × 10−3 + j2π(103 × 0.1 × 10−6) = √ 20 + j2π( 0.25 ) 0.30 −3 + j2π(0.1 × 10−6)
= 169.75 ∠– 25º
Correct Option: A
We have characteristic impedance as follows
Z0 = √ R + jωL G + jωc
R = 20 Ω/ km,
G = 0.3 × 103 s/ km
L = 0.25 mH/ km,
C = 0.1 µF and ω = 103 Hz so we getZ0 = √ 20 + j2π(103 × 0.25 × 10−3) 0.3 × 10−3 + j2π(103 × 0.1 × 10−6) = √ 20 + j2π( 0.25 ) 0.30 −3 + j2π(0.1 × 10−6)
= 169.75 ∠– 25º
- Length of dipole can be calculated from the relationship—
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NA
Correct Option: B
NA