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(D) The conduction current is given by $I_c = \frac{V}{R} = V \sigma A / d$,where $\sigma = 1/\rho$ is the conductivity. The amplitude is $I_{c,0} = \

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$4: 9$

Vedclass · Answer

(D) The conduction current is given by $I_c = \frac{V}{R} = V \sigma A / d$,where $\sigma = 1/ho$ is the conductivity. The amplitude is $I_{c,0} = \frac{V_0}{ho} \frac{A}{d}$.The displacement current is given by $I_d = \epsilon_0 \epsilon_r A \frac{dE}{dt} = \epsilon_0 \epsilon_r A \frac{d}{dt} (V/d) = \frac{\epsilon_0 \epsilon_r A}{d} \omega V_0 \cos(\omega t)$. The amplitude is $I_{d,0} = \frac{\epsilon_0 \epsilon_r A \omega V_0}{d}$.The ratio of the amplitudes is $\frac{I_{d,0}}{I_{c,0}} = \frac{\epsilon_0 \epsilon_r A \omega V_0 / d}{V_0 A / (ho d)} = \epsilon_0 \epsilon_r \omega ho$.Given $\epsilon_r = 80$,$ho = 0.25 \, \Omega m$,$f = 0.4 	imes 10^9 \, Hz$,and $\omega = 2 \pi f = 2 \pi (0.4 	imes 10^9) = 0.8 \pi 	imes 10^9 \, rad/s$.Using $\epsilon_0 = \frac{1}{36 \pi 	imes 10^9} \, F/m$,the ratio is $\frac{1}{36 \pi 	imes 10^9} 	imes 80 	imes (0.8 \pi 	imes 10^9) 	imes 0.25 = \frac{80 	imes 0.8 	imes 0.25}{36} = \frac{16}{36} = \frac{4}{9}$.

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