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A parallel plate capacitor made of circular plate each of radius 10 cm has a capacitance 200 pF. The capacitor is connected to a 200 V supply with an angular frequency of 200 rad/s. What is the peak value of displacement current?

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$2 \sqrt{2} μ A$

$4 \sqrt{2} μ A$

$8 \sqrt{2} μ A$

$4 μ A$

answer is B.

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Detailed Solution

$R M S c u r r e n t, I_{r m s} = \frac{E_{r m s}}{X_{c}} \Rightarrow I_{r m s} = \frac{E_{r m s}}{1 / w C} \Rightarrow I_{r m s} = w C \times E_{r m s} \Rightarrow I_{r m s} = 200 \times 2 \times 10^{- 10} \times 200 = 8 μ A$

Peak value of displacement current is $I_{o} = \sqrt{2} I_{r m s} = \sqrt{2} \times 8 \times 10^{- 6} A = 8 \sqrt{2} μ A$

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