i. Consider a circuit containing a resistor of resistance R, a inductor of inductance L and capacitor of capacitance C connected across an alternating voltage source (Figure). The applied alternating voltage is given by the equation.
`v=V_(m)sinomegat`
ii. Let i be the resulting circuit current in the circuit at that instant. As a result, the voltage is developed across R, L and C.
iii. We know that voltage across R`(V_(R))` is in phase with i, voltage across L`(V_(L))` leads i by `(pi)/(2) and voltage across C`(V_(c))` lags i by `(pi)/(2)`
iv. The phasor diagram is drawn with current as the reference phasor. The current is represented by the phasor `vec(OI),V_(R)"by"vec(OA),V_(L)"by"vec(OB),V_(c)"by"vec(OC)" "" as "`
shown in Figure.
v. The length of these phasors are `OI=I_(m),OA=I_(m)R,OB=I_(m)X_(L),OC=I_(m)X_(c)` The circuit is either effectively inductive or capacitve or resistive that depends on the value of `V_(L)orV_(C).` Let us assume that `V_(L)gtV_(C)` so that net voltage drop across L-C combination is `V_(L)-V_(C)` which is represented by a phasor AD
vi. By parallelogram law, the diagonal `vec(OE)` gives the resultant voltage v of `V_(R)and(V_(L)-V_(C))`and its length OE is equal to `V_(m).` Therefore,
`V_(m)^(2)=V_(R)^(2)+(V_(L)-V_(C))^(2)`
`=sqrt((I_(m)R)^(2)+(I_(m)X_(L)-I_(m)X_(c))^(2))`
`=I_(m)sqrt(R^(2)+(X_(L)-X_(C))^(2))`
`or I_(m)=(V_(m))/(sqrt(R^(2)+(X_(L)-X_(C))^(2)))`
`or I_(m)=(V_(m))/(Z)`
where `Z=sqrt(R^(2)+(X_(L)-X_C)^(2))`
vii. Z is called impedance of the circuit which refers to the effective opposition to the circuit current by the series RLC circuit. The voltage triangle and impedance triangle are given in the Figure.
viii. From phasor diagram, the phase angle between v and i is found out from the following relation
`tanphi=(V_(L)-V_(C))/(V_(R))=(X_(L)-V_(C))/(R)`
Specia cases :
i. If `X_(L)gtX_(C),(X_(L)-X_(C))` is positive and phase angle `phi` is also positive. It means that the applied voltage leads the current by `phi` ( or current lags behind voltage by `phi`). The circuit is inductive.
`therefore v=V_(m)sinomega,i=I_(m)sin(omegat-phi)`
ii. If `X_(L)ltX_(C),(X_(L)-X_(C))` is negative and `phi` is also negative. Therefore current leads voltage by `phi` and the circuit is capacitive.
`thereforev=V_(m)sinomega,i==I_(m)sin(omega+phi)`
iii. If `=X_(L)ltX_(C),phi` is zero. Therefore current and voltage are in the same phase and the circuit is resistive
`thereforev=V_(m)sinomegat,i=I_(m)sinomegat.`
