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A material has conductivity of 10–2 mho/m and a relative permittivity of 4 the frequency at which conduction current in the medium is equal to the displacement current is—
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- 45 MHz
- 90 MHz
- 450 MHz
- 900 MHz
Correct Option: A
As hints are provided in the synopsis that displacement current is the rate of change of electric flux density, so we get
| JC = σ E, JD = | ⇒ ωD = ω ε0εr E | |
| ∂t |
so when both currents are equal
JC = JD ⇒ σE = ω ε0 εr E
| ω = | |
| ε0 εr |
| f = | |
| 2π × 8.85 × 10–12 × 4 |
= 4.49 × 1010–3 × 1010 = 45 MHz
