In a circuit, an inductor (L), capacitor (C) and resistor (R) are connected in parallel across a source of emf given by `epsilon=epsilon_(0) sin omega t`. Find the current through the mains and draw a phasor diagram.

Explanation

As the three elements are connected in parallel, voltage drop across each is same.
Instantaneous current in resistor is `i_(R)=(epsilon_(0))/(R) sin omega t`
Instantaneous current in inductor is `i_(L) = (epsilon_(0))/(X_(L)) sin (omega t- (pi)/(2))`
Instantaneous current in capacitor is `i_(C) = (epsilon_(0))/(X_(C))sin (omega t + (pi)/(2))`
`:. ` Current through mains is `i=(epsilon_(0))/(R) sin omega t + (epsilon_(0))/(X_(L)) sin (omega t - (pi) / (2)) + (epsilon_(0))/(X_(C)) sin (omega t + (pi)/(2))`
or `i=epsilon_(0)[(1)/(R) sin omega t + ((1)/(X_(C))-(1)/(X_(L)))cos omega t]`
Let `(1)/(R) = (1)/(Z) cos phi and (1)/(X_(C))-(1)/(X_(L))=(1)/(Z) sin phi`
`rArr i = (epsilon_(0))/(Z) [ sin omega t cos phi + cos omega t sin phi] = (epsilon_(0))/(Z) sin (omega t + phi)`


As `(1)/(R) = (1) / (Z) cos phi , (1)/(X_(C))-(1)/(X_(L))=((1)/(Z))sin phi`
`rArr (1)/(R^(2))+((1)/(X_(C))-(1)/(X_(L)))^(2)=(1)/(Z^(2))`
`rArr (1)/(Z)=sqrt((1)/(R^(2))+((1)/(X_(C))-(1)/(X_(L)))^(2))`
Phase difference is `tan phi = ((1)/(X_(C))-(1)/(X_(L)))/((1)/(R))` (which is positive is `X_(C) lt X_(L)`)
`i_(0)=(epsilon_(0))/(Z) = sqrt(i_(R)^(2)+ (i_(C)-i_(L))^(2))`