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In an ac circuit an alternating voltage ...

In an ac circuit an alternating voltage `e = 200 sqrt(2) sin 100t` volts is connected to a capacitor of capacity 1 `muF.` The rms.value of the current in the circuit is

Text Solution

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Comparing `E=200sqrt2 sin (100 t)` with `E=E_(0) sin omegat` we find that,
`E_(0)=200sqrt2V` and `omega=100`(rad/s)
So,`X_(C)=1/(omegaC)=1/(100xx10^(-6))=10^(4)Omega`
And as `ac` instruments reads `rms` value, that reading of ammeter will be
`I_(rms)=E_(rms)/X_(C)=E_(0)/(sqrt2X_(C))["as"E_(rms)=E_(0)/sqrt2]`
i.e. `I_(rms)=(200sqrt2)/(sqrt2xx10^(4))=20mA`
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