Seawater at a frequency f = 9 × 102 Hz, has permittivity ∈ = 80∈0 and resistivity ρ = 0.25 Ωm. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source V(t) = V0 sin(2πft). Then the conduction current density becomes 10x times the displacement current density after time t= 1800. The value of x is __________.
(Given : 1/4𝜋∈0 = 9 × 109Nm2C-2)
Answer: (6)
Given:-
f = 9 × 102 Hz
∈=∈0∈r
∈=80∈0
So, ∈r=80
ρ = 0.25 Ωm
V(t) = V0 sin (2πft)
Displacement current, Id = dq/dt=(cdv/dt)
\(\begin{array}{l}I_{d}=\frac{\epsilon _{0}\epsilon _{r}A}{d}\frac{d}{dt}(v_{0}sin(2\pi ft))\end{array} \)
\(\begin{array}{l}I_{d}=\frac{\epsilon _{0}\epsilon _{r}A}{d}V_{0}(2\pi f)cos(2\pi ft)—–(1)\end{array} \)
Where d is the distance between plates & conduction current Ic = V/R
\(\begin{array}{l}I_{c}= \frac{V_{0}sin(2\pi ft)}{\rho \frac{d}{A}}=\frac{AV_{0}sin(2\pi ft)}{\rho d}——(2)\end{array} \)
Divide equation (1) and (2)
\(\begin{array}{l}\frac{I_{d}}{I_{c}}= \epsilon _{0}\epsilon _{r}2\pi f(\rho )cot(2\pi ft)\end{array} \)
\(\begin{array}{l}\frac{I_{d}}{I_{c}}= \frac{1}{4\pi \times 9\times 10^{9}}\times 80\times 2\pi \times 9\times 10^{2}\times (0.25)\times cot(2\pi \times9\times 10^{2}\times \frac{1}{800} )\end{array} \)
\(\begin{array}{l}= \frac{10^{3}}{10^{9}}(cot(\frac{9\pi }{4}))\end{array} \)
\(\begin{array}{l}= \frac{10^{3}}{10^{9}}\end{array} \)
\(\begin{array}{l}\frac{I_{d}}{I_{c}}=\frac{1}{10^{6}}\end{array} \)
Ic = 106 Id
So x = 6
