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How does the capacitor acts as an open circuit at high frequency?

I didn't get.

Hi Nishanth, the capacitor acts as an open circuits because it's capacitive reactance (Something like It's resistence in AC) depends of the frequency of the signal in AC: \[X_{c} = \frac{1}{w{\cdot}C}=\frac{1}{2{\cdot}{\pi}{\cdot}f{\cdot}C}\]. As the frequency gets high in the previous equation also the admitance of the capacitor gets high.

Best regards
 

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