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

`X_(L)=2pifL=2pixx50xx2/pi=200Omega`
`X_(C)=1/(2pi50 100/pixx10^(-5))=100 Omega`
The reactance of the circuit `X=X_(L)-X_(C)=200-100=100 Omega`
Since `X_(1)gt X_(2)` the circuit is called inductive.
(b)impedance of the circuit `Z=sqrt(R^(2)+X^(2))=sqrt(100^(2)-100^(2))=100sqrt2 Omega`
( c)the current `i_(max)=V_(rms)/Z=200/(100sqrt2)=sqrt2A`
(d)reading 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.