An alternating voltage `E=200sqrt(2)sin(100t)` is connected to a `1` microfarad capacitor through an AC ammeter. The reading of the ammeter shall be

Explanation

Consider the purely capacitity circuit as shown below.

Given, V`=200sqrt(2)sin(100t)`…………………(i)
Comparing it with general equation, we get
`V=V_(0)sin(omegat)`………………(ii)
`V_(0) = 200sqrt(2)V, omega=100rads^(-1)`
`therefore` Rectance `X_(C) = 1/(omegaC) = (1/(100 xx 10^(-6)) = 10^(4)Omega`
Peak value of current, `I_(0) = (V_(0)/X_(c)) = (200sqrt(2)/10^(4)) = 2sqrt(2) xx 10^(-2)`A rms value of current
`I_(rms) = I_(0)/sqrt(2) = (2sqrt(2) xx 10^(-2))/(sqrt(2))` = 20 xx 10^(-3) A`= 20 mA
Reading of the ammeter will be same as `I_(rms)`= 20 mA.