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At 50 MHz a lossy dielectric material is characterized by μ = 2.1μo ,ε = 3.6 εo and σ = 0.08 S/m. The electric field is Es = 6e-jγx uz V/m
Que: The impedance η is
- a)101.4Ω
- b)167.4 Ω
- c)98.3 Ω
- d)67.3 Ω
Correct answer is option 'A'. Can you explain this answer?
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At 50 MHz a lossy dielectric material is characterized byμ = 2.1&...
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At 50 MHz a lossy dielectric material is characterized byμ = 2.1&...
Impedance Calculation in Lossy Dielectric Material
Given:
Frequency (f) = 50 MHz = 50 x 10^6 Hz
Loss tangent (δ) = 2.1°
Relative permittivity (εr) = 3.6
Conductivity (σ) = 0.08 S/m
Electric field (Es) = 6e^(-jxuz) V/m
Impedance Calculation:
The impedance (Z) of a lossy dielectric material can be calculated using the following formula:
Z = sqrt((jωμ) / (σ + jωε))
Where:
ω = 2πf (Angular frequency)
μ = Permeability of free space = 4π x 10^(-7) H/m
σ = Conductivity of the material
ε = εrε0 (Permittivity of the material)
εr = Relative permittivity
ε0 = Permittivity of free space = 8.854 x 10^(-12) F/m
1. Angular Frequency Calculation:
ω = 2πf
= 2π x 50 x 10^6
= 100π x 10^6 rad/s
2. Permeability Calculation:
μ = 4π x 10^(-7) H/m
3. Permittivity Calculation:
ε = εrε0
= 3.6 x 8.854 x 10^(-12)
= 31.8744 x 10^(-12) F/m
4. Impedance Calculation:
Z = sqrt((jωμ) / (σ + jωε))
First, let's calculate the numerator:
Numerator = jωμ
= (j)(100π x 10^6)(4π x 10^(-7))
= 400π^2 x 10^(-1) j
Next, let's calculate the denominator:
Denominator = (σ + jωε)
= (0.08 + j)(100π x 10^6)(31.8744 x 10^(-12))
= (8π x 10^6 + j)(31.8744 x 10^(-6))
= 254.9952π x 10^(-6) + j x 1008π x 10^(-6)
Now, let's simplify the impedance formula:
Z = sqrt((jωμ) / (σ + jωε))
= sqrt((400π^2 x 10^(-1) j) / (254.9952π x 10^(-6) + j x 1008π x 10^(-6)))
To simplify further, let's rationalize the denominator:
Z = sqrt((400π^2 x 10^(-1) j) / ((254.9952π x 10^(-6) + j x 1008π x 10^(-6))) x ((254.9952π x 10^(-6) - j x 1008π x 10^(-6))) / ((254.9952π x 10^(-6) - j x 1008π x 10^(-6))))
Simplifying the denominator:
Z = sqrt((400π^2
Given:
Frequency (f) = 50 MHz = 50 x 10^6 Hz
Loss tangent (δ) = 2.1°
Relative permittivity (εr) = 3.6
Conductivity (σ) = 0.08 S/m
Electric field (Es) = 6e^(-jxuz) V/m
Impedance Calculation:
The impedance (Z) of a lossy dielectric material can be calculated using the following formula:
Z = sqrt((jωμ) / (σ + jωε))
Where:
ω = 2πf (Angular frequency)
μ = Permeability of free space = 4π x 10^(-7) H/m
σ = Conductivity of the material
ε = εrε0 (Permittivity of the material)
εr = Relative permittivity
ε0 = Permittivity of free space = 8.854 x 10^(-12) F/m
1. Angular Frequency Calculation:
ω = 2πf
= 2π x 50 x 10^6
= 100π x 10^6 rad/s
2. Permeability Calculation:
μ = 4π x 10^(-7) H/m
3. Permittivity Calculation:
ε = εrε0
= 3.6 x 8.854 x 10^(-12)
= 31.8744 x 10^(-12) F/m
4. Impedance Calculation:
Z = sqrt((jωμ) / (σ + jωε))
First, let's calculate the numerator:
Numerator = jωμ
= (j)(100π x 10^6)(4π x 10^(-7))
= 400π^2 x 10^(-1) j
Next, let's calculate the denominator:
Denominator = (σ + jωε)
= (0.08 + j)(100π x 10^6)(31.8744 x 10^(-12))
= (8π x 10^6 + j)(31.8744 x 10^(-6))
= 254.9952π x 10^(-6) + j x 1008π x 10^(-6)
Now, let's simplify the impedance formula:
Z = sqrt((jωμ) / (σ + jωε))
= sqrt((400π^2 x 10^(-1) j) / (254.9952π x 10^(-6) + j x 1008π x 10^(-6)))
To simplify further, let's rationalize the denominator:
Z = sqrt((400π^2 x 10^(-1) j) / ((254.9952π x 10^(-6) + j x 1008π x 10^(-6))) x ((254.9952π x 10^(-6) - j x 1008π x 10^(-6))) / ((254.9952π x 10^(-6) - j x 1008π x 10^(-6))))
Simplifying the denominator:
Z = sqrt((400π^2
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At 50 MHz a lossy dielectric material is characterized byμ = 2.1μo,ε = 3.6εoandσ = 0.08 S/m.The electric field is Es= 6e-jγxuzV/mQue: The impedance η isa)101.4Ωb)167.4 Ωc)98.3 Ωd)67.3 ΩCorrect answer is option 'A'. Can you explain this answer?
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At 50 MHz a lossy dielectric material is characterized byμ = 2.1μo,ε = 3.6εoandσ = 0.08 S/m.The electric field is Es= 6e-jγxuzV/mQue: The impedance η isa)101.4Ωb)167.4 Ωc)98.3 Ωd)67.3 ΩCorrect answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about At 50 MHz a lossy dielectric material is characterized byμ = 2.1μo,ε = 3.6εoandσ = 0.08 S/m.The electric field is Es= 6e-jγxuzV/mQue: The impedance η isa)101.4Ωb)167.4 Ωc)98.3 Ωd)67.3 ΩCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 50 MHz a lossy dielectric material is characterized byμ = 2.1μo,ε = 3.6εoandσ = 0.08 S/m.The electric field is Es= 6e-jγxuzV/mQue: The impedance η isa)101.4Ωb)167.4 Ωc)98.3 Ωd)67.3 ΩCorrect answer is option 'A'. Can you explain this answer?.
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