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In the circuit shown in figureure the `AC` source gives a voltage `V=20cos(2000t)`. Neglecting source resistance, the voltmeter and and ammeter readings will be
image.
A. `0V, 0.47 A`
B. `1.68V,0.47 A`
C. `0V,1.4A`
D. `5.6 V, 1.4 A`

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Best answer
Correct Answer - D
`Z=sqrt((R)^(2)+(X_(L)-X_(C))^(2))`,
`R=10 Omega, X_(L)=omegaL=2000xx5xx10^(-3)=10 Omega`
`X_(C)=1/(omegaC)=1/(2000xx50xx10^(-6))=10 Omega i.e., Z=10 Omega`
Maximum current `i_(0)=(V_(0))/Z=20/10=2A`
Hence `i_(rms)=2/(sqrt(2))=1.4A`
and `V_(rms)=4xx1.41=5.46 V`

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