151. A uniform plane wave travelling in air is incident on the plane boundary between air and another dielectric medium with ∈_r = 4. The reflection co-efficient for the normal incidence is—

1. 1. 0

   
   
   

   2. 0.5 ∠180º

   
   
   

   3. 0.333 ∠0º

   
   
   

   4. 0.333 ∠180º

   
   
   

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1. View Hint View Answer Discuss in Forum

   Reflection coefficient in terms of dielectric constant is given as
   

   Γ =	√∈r2 - √∈r1	=	√4 - √1
   	√∈r2 + √∈r1		√4 + √1

   
   

   =	2 - 1	=	1	= 0.333 ∠0°
   	2 + 1		3

   
   Hence alternative (D) is the correct choice.

   Correct Option: D

   Reflection coefficient in terms of dielectric constant is given as
   

   Γ =	√∈r2 - √∈r1	=	√4 - √1
   	√∈r2 + √∈r1		√4 + √1

   
   

   =	2 - 1	=	1	= 0.333 ∠0°
   	2 + 1		3

   
   Hence alternative (D) is the correct choice.

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152. A stub having load impedance Z_L and stub length λ/ 2 then input impedance Z_in is—

1. 1. Z_L

   
   
   

   2. ZL

      2

   
   
   

   3. √Z_L

   
   
   

   4. Insufficient data
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Zin = Z0		ZL + j Z0 tanβl
   		Z0 + j ZL tanβl

   
   

   Zin = Z0		ZL + j Z0 tan π
   		Z0 + j ZL tan π

   
   

   βl =	2π	,	λ	= π
   	λ		2

   
   ∴ l = λ/2
   

   Zin =	Z0. ZL	, Zin = ZL
   	Z0

   Correct Option: A

   Zin = Z0		ZL + j Z0 tanβl
   		Z0 + j ZL tanβl

   
   

   Zin = Z0		ZL + j Z0 tan π
   		Z0 + j ZL tan π

   
   

   βl =	2π	,	λ	= π
   	λ		2

   
   ∴ l = λ/2
   

   Zin =	Z0. ZL	, Zin = ZL
   	Z0

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153. A short circuited stub is shunt connected to a T.L. as shown. If Z_o = 50 Ω, the admittance Y seen at the function of stub and the T.L. is—
     

1. 1. (0.01 – j 0.02) mho

   
   
   

   2. (0.02 – j 0.01) mho

   
   
   

   3. (0.04 – j 0.02) mho

   
   
   

   4. (0.01 – j 0) mho

   
   
   

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1. View Hint View Answer Discuss in Forum

   When
   

   l =	λ
   	8

   
   For short circuit stub
   

   Zin = Z0		ZL + j Z0 tan βI
   		Z0 + j ZL tan βI

   
   

   = Z0		0 + j 50 tan π/4
   		Z0 + j 0 tan π/4

   
   

   =	Z0[j 50]		∵ 2π/λ , λ/8 = π/4
   	Z0		ZL = 0

   
   = j 50
   or
   

   Yin =	1	= - 0.02
   	√50

   
   when
   I = λ/2 then βl = 2π/ λ · λ/2 = π
   

   Zin2 = Z0		ZL + j Z0 tan π
   		Z0 + j ZL tan π

   
   

   = Z0		ZL + j Z0 0
   		Z0 + j ZL 0

   
   Z_L = 100
   or
   Y_in2 = 1/100 = 0.01
   so resultant admittance Y = Y_in1 + Y_in2 = 0.01 – j 0.02 mho.
   

   Correct Option: A

   When
   

   l =	λ
   	8

   
   For short circuit stub
   

   Zin = Z0		ZL + j Z0 tan βI
   		Z0 + j ZL tan βI

   
   

   = Z0		0 + j 50 tan π/4
   		Z0 + j 0 tan π/4

   
   

   =	Z0[j 50]		∵ 2π/λ , λ/8 = π/4
   	Z0		ZL = 0

   
   = j 50
   or
   

   Yin =	1	= - 0.02
   	√50

   
   when
   I = λ/2 then βl = 2π/ λ · λ/2 = π
   

   Zin2 = Z0		ZL + j Z0 tan π
   		Z0 + j ZL tan π

   
   

   = Z0		ZL + j Z0 0
   		Z0 + j ZL 0

   
   Z_L = 100
   or
   Y_in2 = 1/100 = 0.01
   so resultant admittance Y = Y_in1 + Y_in2 = 0.01 – j 0.02 mho.
   

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154. Distilled water at 25ºC is characterized by σ = 1.7 × 10^–4 mho/m, ε = 78 ∈_o at a frequency 3 GHz. Its loss tangent is—

1. 1. 1.345 × 10^{– 5} degree

   
   
   

   2. 1.345 × 10^{– 1} degree

   
   
   

   3. 1.345 × 10^{– 2} degree

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Loss tangent = tan θ = σ ω∈
   

   =	1.7 × 10–4
   	2π × 3 × 109 × 78 × 8.6 × 10–12

   (˙.˙ ∈₀ = 8.6 × 10^–12)
   = 1.345 × 10^–5 degree.

   Correct Option: A

   Loss tangent = tan θ = σ ω∈
   

   =	1.7 × 10–4
   	2π × 3 × 109 × 78 × 8.6 × 10–12

   (˙.˙ ∈₀ = 8.6 × 10^–12)
   = 1.345 × 10^–5 degree.

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155. 8-point charges of 1 nc each are located at the corners of a cube in free space that is 1 m on a side, then ^→E at the centre is—

1. 1. 0 V/m

   
   
   

   2. 5 V/m

   
   
   

   3. 4 V/m

   
   
   

   4. – 5 V/m

   
   
   

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1. View Hint View Answer Discuss in Forum

   ^→E at the centre is always zero.
   

   Correct Option: A

   ^→E at the centre is always zero.
   

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