Displacement current

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Question

A Parallel plate condenser of capacity 100 pF is connected to 230 V of AC supply of 300 rad / sec frequency. The rms value of displacement current

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Solution

$\begin{array}{rcl} i_{d} & = & \in_{0} \frac{dϕ}{dt} = \frac{ε_{0} A}{d} (\frac{dV}{dt}) = C \frac{dV}{dt} = C \frac{d}{dt} (V_{0} sinωt) \\ i_{d} & = & {CωV}_{0} cosωt = i_{0} cosωt \\ v_{rms} & = & 230 V \\ i_{rms} & = & \frac{i_{0}}{\sqrt{2}} = \frac{{CωV}_{0}}{\sqrt{2}} (∴ v_{rms} = \frac{V_{0}}{\sqrt{2}}) \\ = & \frac{100 x 10^{- 12} x 300 x 230 \sqrt{2}}{\sqrt{2}} = 6.9 μA \end{array}$

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Displacement current

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A Parallel plate condenser of capacity 100 pF is connected to 230 V of AC supply of 300 rad / sec frequency. The rms value of displacement current

id =∈0dϕdt=ε0AddVdt=CdVdt=CddtV0 sinωtid =CωV0 cosωt=i0 cosωt vrms =230 Virms =i02=CωV02 ∴vrms=V02 =100 x 10-12 x 300 x 23022=6.9 μA

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