In the circuit shown in the figure, find:
(a) the reactance of the circuit
(b) impedance of the circuit
(c )the current
(d) readings of the ideal AC voltmeters(these are hot wire instruments and read rms values)

Explanation

`(a)X_(L)=2pifL=2pixx50xx(2)/(pi)=200Omega`
`X_(C)=(1)/(2pi50(100)/(pi)xx10^(-6))=100Omega`
`therefore` The reactance of the circuit `X=X_(L)-X_(C)=200-100=100Omega`
Since, `X_(L)-X_(C)` the circuit is called inductive.
(b) impedence of the circuit `Z=sqrt(R^(2)+X^(2))=sqrt(100^(2)+100^(2))=100Omega`
(c ) the current `I_("rms")=(V_("rms"))/(Z)=(200)/(100sqrt2)=sqrt2A`
(d) readings of the ideal voltmeter
`V_(1)=I_("rms")X_(L)=200sqrt2"Volt"`
`V_(2)=I_("rms")R=100sqrt2"Volt"`
`V_(3)=I_("rms")X_(c)=100sqrt2"Volt"`
`V_(4)=I_("rms")sqrt(R^(2)+X_(L)^(2))=100sqrt10"Volt"`
`V_(5)=I_("rms")Z=200"volt"` which also happens to be the voltage of source.