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In the series curcuit of fig, suppose R=...

In the series curcuit of fig, suppose `R=300 Omega,L=60mH,C=0.50(mu)F`, source amplitude is `E_(0)=50V and omega=10000 rad s^(-1)`. Find the reactances `X_(L) and X_(C)`, the impedance Z and the current amplitude `I_(0)`.
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Text Solution

Verified by Experts

The inducitive and capacitive reactances are
`X_(L)=(1)/(omega C)=(1)/((10,000 rad s^(-1))(0.50xx10^(-6)F))=200 Omega`
The impedances Z of the circuit is
`Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))`
`=sqrt((300 Omega)^(2)+(600 Omega-200 Omega)^(2))=500 Omega`
With source voltage amplitude `E_(0)=50V`, The current amplitude
is `I_(0)=(E_(0))/(Z)=(50V)/(500Omega)=0.10A`
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