Electromagnetic theory miscellaneous
- A 300 Ω line is terminated in a load impedance of 100 + j 200 Ω the voltage reflection coefficient is—
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By basic formulae of reflection coefficient, we get
Γ = ZL − Zc ZL + Zc = 100 + j 200 − 300 100 + j 200 + 300 = j 200 − 200 j 200 + 400 = j 2 − 2 j 2 + 4
= 0.6325 ∠108°Correct Option: A
By basic formulae of reflection coefficient, we get
Γ = ZL − Zc ZL + Zc = 100 + j 200 − 300 100 + j 200 + 300 = j 200 − 200 j 200 + 400 = j 2 − 2 j 2 + 4
= 0.6325 ∠108°
- A metal sphere with 1 m radius and a surface charge density of 10 coulomb/m2 is enclosed in a cube of 10 m side. The total outward electric displacement normal to the surface of the cube is—
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We have total charge enclosed by cube
qtotal = A. σ = 4 σr2 × 10 = 40 σπ Gauss's theorem
Total outward flux = Total charge enclosed
∫ D·dS = qtotal = 40∏
(given that d = 10 m)Correct Option: D
We have total charge enclosed by cube
qtotal = A. σ = 4 σr2 × 10 = 40 σπ Gauss's theorem
Total outward flux = Total charge enclosed
∫ D·dS = qtotal = 40∏
(given that d = 10 m)
- A uniform plane wave in air is normally incident on infinitely thick slab, if the refractive index of the glass slab is 1.5 then the percentage of incident power that is reflected from the air glass interface is—
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As we know that ratio of the reflected electric field on a slab of intrinsic impedance η2 and η1 to the incidence electric field is given by following relation
Γ = Er = η2 − η1 (for normal incidence) Ei η2 + η1 
⇒ √ε₁ - √ε₂ √ε₁ + √ε₂
where √ε₂ = refractive index of medium of slab so we get|Γ| = Er = η1 − η2 = 1.5 − 1 = 1 Ei η1 + η2 1.5 + 1 5
Power reflected = | Γ |2 PinciPref = 
1 
2 Pinc = 1 Pinc 5 25 Pref = 1 × 100 = 4 % 25 Correct Option: B
As we know that ratio of the reflected electric field on a slab of intrinsic impedance η2 and η1 to the incidence electric field is given by following relation
Γ = Er = η2 − η1 (for normal incidence) Ei η2 + η1 ⇒ √ε₁ - √ε₂ √ε₁ + √ε₂
where √ε₂ = refractive index of medium of slab so we get|Γ| = Er = η1 − η2 = 1.5 − 1 = 1 Ei η1 + η2 1.5 + 1 5
Power reflected = | Γ |2 PinciPref = 1 2 Pinc = 1 Pinc 5 25 Pref = 1 × 100 = 4 % 25
- Critical frequency of ionospheric layer is 10 MHz what is the maximum launching angle from the horizon for which 20 MHz wave will be reflected by the layer?
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Critical frequency 'fc' is the frequency at which wave could be reflected from the ionosphere and maximum usable frequency is that frequency at which the wave could be successfully used by receiver, so we get
fmuf = fc sec i
⇒ 20 MHz = 10 MHz sec i
sec i = 2 ⇒ i = 60º
however, the angle from horizontal, we get β = 30º (i.e. 90º – 60º = 30º)Correct Option: B
Critical frequency 'fc' is the frequency at which wave could be reflected from the ionosphere and maximum usable frequency is that frequency at which the wave could be successfully used by receiver, so we get
fmuf = fc sec i
⇒ 20 MHz = 10 MHz sec i
sec i = 2 ⇒ i = 60º
however, the angle from horizontal, we get β = 30º (i.e. 90º – 60º = 30º)
- A km long microwave link uses two antennas each having 30 dB gain. If the power transmitted by one antenna is 1 W at 3 GHz. Power received by the other antenna is approximately—
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As we know from basic equation of transmitter that
PR = (GT)dB + (GR)dB – (32.5 + 20 log r + 20 log f) PT dB
where r in km and f in MHz so we get by putting the valuesPR = 30 + 30 – (32.5 + 0 + 60 + 9.542) PT dB
= 60 - (102.4)log PR = −4.205 PT
∴ PR = PT × 63.27 × 10−6
= 63.27 μWCorrect Option: C
As we know from basic equation of transmitter that
PR = (GT)dB + (GR)dB – (32.5 + 20 log r + 20 log f) PT dB
where r in km and f in MHz so we get by putting the valuesPR = 30 + 30 – (32.5 + 0 + 60 + 9.542) PT dB
= 60 - (102.4)log PR = −4.205 PT
∴ PR = PT × 63.27 × 10−6
= 63.27 μW
