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AC voltage source (V, omega) is applied ...

AC voltage source `(V, omega)` is applied across a parallel LC circuit as shown in figure. Find the impedance of the circuit and phase of current.

Text Solution

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Current through the resistor, `I_(R) = V//R`, in phase with voltage
Current through the inductor, `I_(L)=V/X_(L)`, lags voltage by `pi//2`
Current through the capacitor, `I_(C)=V//V_(C)`, leads voltage by `pi//2`

Assuming
Resultant current, I`=(I_(R)^(2)+(I_(C)-I_(L))^(2)]^(1//2)`
`rArr V/Z = [(V/R)^(2)+(V/X_(C)-V/(X_(L))^(2))]^(1//2)`
`rArr 1/Z = [1/R^(2)+(1/X_(C)-1/X_(L))^(2)]^(1//2)`
Impedance, `Z=[1/R^(2)+(omegaC-1/(omegaL))^(2)]^(-1//2)`
Phase difference, `tanphi = (I_(C)-I_(L))/I_(R) = ((1/X_(C))-1/(X_(L)))/(1//R)`
`phi = tan^(-1)[R(omegaC-1/(omegaL))]`
This means current leads the voltage.
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