An alternating voltage `E=200sqrt(2) sin (100 t)V` is connected to a `1 (mu)F` capaacitor through an ac ammeter (it reads rms value). What will be the reading of he ammeter?

An alternating voltage `E=200sqrt(2) sin (100 t)V` is connected to a `

1.	An alternating voltage `E=200sqrt(2) sin (100 t)V` is connected to a `1 (mu)F` capaacitor through an ac ammeter (it reads rms value). What will be the reading of he ammeter?
Answer» Correct Answer - 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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An alternating voltage `E=200sqrt(2) sin (100 t)V` is connected to a `1 (mu)F` capaacitor through an ac ammeter (it reads rms value). What will be the reading of he ammeter?

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