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, the current amplitude `I_(0)`, the phase angle `phi`, and the voltage amplitude across each circuit element.
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Explanation

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`
the phase angle `phi` is
`phi=tan^(-1)(X_(C)-X_(L))/(R )=tan^(-1)(-(400 Omega)/(300 Omega))=-53^(@)`
The voltage amplitudes `V_(R0), L_(L0), and V_(C0)` across the resister, inductor, and capacitor, respectively are
`V_(R0)=I_(0)R=(0.10A)(300 Omega)=30V`
`V_(L0)=I_(0)X_(L)=(0.10A)(200 Omega)=20V`
Note that `X_(L)gtX_(C)` and hebce the voltage amplitude across the inductor is greater than that across the capacitor and `phi` is negative. The value `phi=-53^(@)` means that the voltage leads the currents by `53^(@)`, this is like the situation shown in fig.
Note that the source voltage amplitude across the separate circuit elements (that is , 50V pi 30V+60V+0V`)
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