JEE PYQ: Current Electricity Question 9
Question 9 - 2021 (17 Mar 2021 Shift 2)
Seawater at a frequency $f = 9 \times 10^2$ Hz, has permittivity $\varepsilon = 80\varepsilon_0$ 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 $V(t) = V_0 \sin(2\pi ft)$. Then the conduction current density becomes $10^x$ times the displacement current density after time $t = \frac{1}{800}$ s. The value of $x$ is ____. (Given: $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9,\text{Nm}^2\text{C}^{-2}$)
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Answer: 6
Solution
$J_c = \frac{E}{\rho} = \frac{V}{\rho d}$. $J_d = \frac{\varepsilon}{d} \frac{dV_c}{dt} = \frac{\varepsilon}{d} V_0 (2\pi f) \cos 2\pi ft$. $\frac{J_c}{J_d} = \frac{V_0 \sin 2\pi ft}{\rho d} \times \frac{d}{80\varepsilon_0 V_0 (2\pi f) \cos 2\pi ft}$. At $t=1/800$: $\tan(2\pi \times 900/800) = 10^x \times \frac{40}{9 \times 10^9} \times 900$. $x = 6$.
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