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An alternating voltage E = 200sqrt 2sin(...

An alternating voltage E = 200`sqrt 2`sin(100t) is connected to a `mu`F capacitor through an AC ammeter. The reading of the ammeter shall be

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The correct Answer is:
20mA
`X_(C ) =(1)/(omega C) =(1)/(100xx10^(-6)) =(10^4) (Omwega)`
As ac instruments read rms value, The reading of ammeter will be,
`I_(rms)=(E_(rms))/(X_C) =(E_0)/(sqrt(2)X_(C ))` `{as E_(rms)=(E_0)/(sqrt(2))}`
i.e. `I_(rms)=(200 sqrt(2))/(sqrt(2)xx10^(4)) = 0.02A = 20mA`.
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