You are given three circuits X, Y and Z when the element X is connected across an a.c. source of a given voltage, the current and voltage are in same phase. When element Y is connected in series with X across the source, voltage is ahead of voltage by `pi//4`. But the current is ahead of voltage by `pi//4`, when Z is connected series with X, across the source. Identify the circuit elements X, Y and Z.
When all the three elements are connected in series across the same source, determine the impedance of the circuit . Draw a plot of current versus the frequency of applied source and mention the significance of this plot.

Explanation

With X, current and voltage are in same phase, therefore, element X must be pure resistor R. With X and Y in series, voltage is ahead of current, therefore, Y must be inductor.
`(tan) (pi)/(4) = (R )/(sqrt(R^(2) + X_(L)^(2))) = 1 :. X_(L) = 0`
With X and Z in series, current is ahead of voltage, therefore Z must be a capacitor.
`(tan) (pi)/(4) = (R )/(sqrt(R^(2) + X_(C )^(2)) )= 1 :. X_(C ) = 0`
When all the three elements are connected in series across the same source.
`Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2)) = R`
It si a series resoance circuit. Current is maximum at resonance frequency `omega_(r) = (1)/(sqrt(LC))`
as show in Fig.