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Question
Derive an expression for phase angle between the applied voltage and current in a series RLC circuit.
Solution
AC circuit containing R, L and C
L(VL) leads I by `π/2` and voltage across C(VC) lags I by `π/2`
Phasor diagram is drawn with the current.
The length of these phasors are
OI = Im, OA = ImR, OB = ImχL;
OC = ImχC
Phasor diagram for a series RLC – circuit when VL > VC
Vm2 = VR2 + (VL – VC)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 χL> χC, (χL – χC) is positive and Φ is also positive.
∴ υ = Vmsin ωt; i = Im(sin(ωt – Φ))
ii) if χL < χC, (χL – χC) is negative and Φ is negative.
∴ υ = Vmsin ωt; i = Im(sin(wt + Φ))
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