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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

Text Solution

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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.
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