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Standard XII

Physics

Charging of Capacitors

On your first...

Question

On your first day as an electrical technician you are asked to determine the leakage current in a capacitor. The capacitor consists of two parallel-plates. The plates have equal and opposite charge $Q$ . The dielectric has dielectric constant $K$ and a resistivity $ρ$ . The plates have area $A$ and separation between the two plates is $d$ . The leakage resistance is :

R = \frac{ρ A}{d}

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R = \frac{ρ d}{A}

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R = \frac{ρ d}{K A}

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R = \frac{ρ d^{2}}{A} K

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Solution

The correct option is C $R = \frac{ρ d}{A}$
The leakage resistant $R = \frac{ρ ϵ}{C^{'}}$
Here, $C^{'} = k C,$ where $k = \frac{ϵ}{ϵ_{0}}$ and $C = \frac{A ϵ_{0}}{d}$
$R = \frac{ρ k ϵ_{0}}{k C} = \frac{ρ d}{A}$

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The correct option is C R=ρdAThe leakage resistant R=ρϵC′Here, C′=kC, where k=ϵϵ0 and C=Aϵ0dR=ρkϵ0kC=ρdA

The correct option is C $R = \frac{ρ d}{A}$
The leakage resistant $R = \frac{ρ ϵ}{C^{'}}$
Here, $C^{'} = k C,$ where $k = \frac{ϵ}{ϵ_{0}}$ and $C = \frac{A ϵ_{0}}{d}$
$R = \frac{ρ k ϵ_{0}}{k C} = \frac{ρ d}{A}$