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A material has a conductivity of 10-2 mho/m and a relative permittivity of 10. The frequency at which the conduction current in the medium, is equal to the displacement current is ______. (Answer should be in MHz)
Correct answer is '18'. Can you explain this answer?
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A material has a conductivity of 10-2 mho/m and a relative permittivi...
Given :σ = 102 mho/m
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εr = 10
Now, magnitude of displacement current density for sinusoidal fields is given as,
And magnitude of conduction current density,
As per given data, equation (i) = (ii),
But 
f = 18 x 106 Hz = 18MHz
Most Upvoted Answer
A material has a conductivity of 10-2 mho/m and a relative permittivi...
Given :σ = 102 mho/m
εr = 10
Now, magnitude of displacement current density for sinusoidal fields is given as,
And magnitude of conduction current density,
As per given data, equation (i) = (ii),
But
f = 18 x 106 Hz = 18MHz
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A material has a conductivity of 10-2 mho/m and a relative permittivi...
Understanding Conductivity and Displacement Current
In electrical engineering, the total current in a medium can be understood as a combination of conduction current and displacement current. The key to solving this problem lies in finding the frequency at which these two currents are equal.
Given Parameters:
- Conductivity (σ) = 10^(-2) mho/m
- Relative Permittivity (εr) = 10
Calculating the Permittivity:
The absolute permittivity (ε) is calculated using the formula:
- ε = ε0 * εr
Where ε0 (permittivity of free space) is approximately 8.854 x 10^(-12) F/m.
Thus:
- ε = 8.854 x 10^(-12) * 10 = 8.854 x 10^(-11) F/m
Finding Angular Frequency:
To determine the frequency at which the conduction current equals the displacement current, we use the relationship:
- σ = ω * ε
Where ω is the angular frequency.
Rearranging gives:
- ω = σ / ε
Substituting the values:
- ω = (10^(-2)) / (8.854 x 10^(-11))
Calculating ω results in:
- ω ≈ 1.13 x 10^8 rad/s
Converting to Frequency:
To convert angular frequency to frequency (f), use:
- f = ω / (2π)
Calculating this gives:
- f ≈ 18.0 MHz
Thus, the frequency at which the conduction current equals the displacement current is approximately 18 MHz.
In electrical engineering, the total current in a medium can be understood as a combination of conduction current and displacement current. The key to solving this problem lies in finding the frequency at which these two currents are equal.
Given Parameters:
- Conductivity (σ) = 10^(-2) mho/m
- Relative Permittivity (εr) = 10
Calculating the Permittivity:
The absolute permittivity (ε) is calculated using the formula:
- ε = ε0 * εr
Where ε0 (permittivity of free space) is approximately 8.854 x 10^(-12) F/m.
Thus:
- ε = 8.854 x 10^(-12) * 10 = 8.854 x 10^(-11) F/m
Finding Angular Frequency:
To determine the frequency at which the conduction current equals the displacement current, we use the relationship:
- σ = ω * ε
Where ω is the angular frequency.
Rearranging gives:
- ω = σ / ε
Substituting the values:
- ω = (10^(-2)) / (8.854 x 10^(-11))
Calculating ω results in:
- ω ≈ 1.13 x 10^8 rad/s
Converting to Frequency:
To convert angular frequency to frequency (f), use:
- f = ω / (2π)
Calculating this gives:
- f ≈ 18.0 MHz
Thus, the frequency at which the conduction current equals the displacement current is approximately 18 MHz.
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A material has a conductivity of 10-2 mho/m and a relative permittivity of 10. The frequency at which the conduction current in the medium, is equal to the displacement current is ______. (Answer should be in MHz)Correct answer is '18'. Can you explain this answer?
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