Question

Seawater at a frequency $f=9\times {10}^2\text{Hz}$, has permittivity $\epsilon =80{\epsilon }_0$ and resistivity $\rho =0.25\Omega \text{m}$. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source $V\left(t\right)=V_0\sin \left(2\pi ft\right)$. Then, the conduction current between the plates becomes ${10}^x$ times the displacement current after time $t=1/800\text{s}$. The value of $x$ (integer only) is

• A. 6
• B. 6.00
• C. 6.0

Answer 1

Answer: (A), (B), (C)

The displacement current is given by

$I_d=\frac{\epsilon A}{d}\frac{dV}{dt}$

$I_d=\frac{\epsilon A}{d}\frac{d\left[V_0\sin \right(2\pi ft\left)\right]}{dt}$

$I_d=\frac{\epsilon A{V}_0\times 2\pi f}{d}\cos \left(2\pi ft\right).......\left(1\right)$

The condcution current between the plates is given by

$I_C=\frac{V}{\rho \frac{d}{A}}(\because R=\frac{\rho d}{A})$

$I_C=\frac{V_{0}A\sin \left(2\pi ft\right)}{\rho d}.......\left(2\right)$

From eqn (1) and (2)

$\frac{I_d}{I_C}=\epsilon \times 2\pi f\times \rho \cot \left(2\pi ft\right)$

Here,
$\epsilon =80{\epsilon }_0;t=1/800\text{s};\rho =0.25\Omega \text{m}$

$f=9\times {10}^2\text{Hz}$

$\frac{I_d}{I_C}=80\times \frac{1}{4\pi \times 9\times {10}^9}\times 2\pi \times 9\times {10}^2\times 0.25\times \cot (2\pi \times 9\times {10}^2\times \frac{1}{800})$

$\frac{I_d}{I_c}={10}^{-6}$

$\Rightarrow I_C={10}^6{I}_d$

$\therefore x=6$