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An AC voltage source E=200sqrt(2), is co...

An AC voltage source `E=200sqrt(2)`, is connected across AC ammeter and capacitor of capacitance `1muF`, the reading of ammeter is `(omega=rad//s)`

A

80 mA

B

40 mA

C

10 mA

D

20 mA

Text Solution

Verified by Experts

The correct Answer is:
D
Impendance of the circuit is
`Z=X_(C)=(1)/(omegaC)=(1)/(100xx1xx10^(-6))`
Thus, `l=(E_(rms))/(Z)=(200)/(10^(4))=20mA`
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