Question

Derive an expression for phase angle between the applied voltage and current in a series RLC circuit.

Answer 1

AC circuit containing R, L and C

L(V_L) leads I by `π/2` and voltage across C(V_C) lags I by `π/2`

Phasor diagram is drawn with the current.

The length of these phasors are

OI = I_m, OA = I_mR, OB = I_mχ_L;

OC = I_mχ_C

Phasor diagram for a series RLC – circuit when V_L > V_C

V_m² = V_R² + (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"/"Z"`

where Z = `sqrt("R"^2 + ("X"_"L" - "X"_"C")^2)` It is impedence of circuit.

From phasor diagram, phase angle between angle between v and i is

`tan phi = ("V"_"L" - "V"_"L")/"V"_"R" = ("X"_"L" - "X"_"C")/"R"`

Special cases:

i) If χ_L> χ_C, (χ_L – χ_C) is positive and Φ is also positive.

∴ υ = V_msin ωt; i = I_m(sin(ωt – Φ))

ii) if χ_L < χ_C, (χ_L – χ_C) is negative and Φ is negative.

∴ υ = V_msin ωt; i = I_m(sin(wt + Φ))