Question

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

Accepted Answer

Let series RLC circuit is connected with AC source having Voltage, $V = V_{m} \sin ω t$

Voltage across resistor, $V_{R} = i R$
Voltage across inductor, $V_{L} = L \frac{d i}{d t}$
Voltage across capacitor, $V_{C} = \frac{q}{C}$
Applying Kirchhoff's law: $V_{R} + V_{L} + V_{C} = V$
$\Rightarrow i R + L \frac{d i}{d t} + \frac{q}{C} = V_{m} \sin ω t$
Differentiating with respect to time:
$\Rightarrow \frac{d i}{d t} R + L \frac{d^{2} i}{d t^{2}} + \frac{i}{C} = V_{m} ω \cos ω t$ .........(1)
Assume that, $i = A \sin (ω t + Φ)$ , where $A$ is the peak current and $Φ$ is the phase difference between voltage and current.
$\Rightarrow \frac{d i}{d t} = A ω \cos (ω t + Φ)$ and $\frac{d^{2} i}{d t^{2}} = - A ω^{2} \sin (ω t + Φ)$

Hence, equation $(1)$ becomes,
$\Rightarrow A ω \cos (\cot + Φ) R - L A ω^{2} \sin (ω t + Φ) + \frac{A}{C} \sin (ω t + Φ) = V_{m} ω \cos ω t$ $\Rightarrow A ω R \cos (ω t + Φ) + (\frac{A}{C} - L A ω^{2}) \sin (ω t + Φ) = V_{m} ω \cos ω t$
$\Rightarrow A R \cot (ω t + Φ) + A [\frac{1}{ω C} - ω L] \sin (ω t + Φ) = V_{m} \cos ω t$
$\Rightarrow A \sqrt{R^{2} + {(\frac{1}{ω C} - ω L)}^{2}} [\frac{R}{\sqrt{R^{2} + {(\frac{1}{ω C} - ω L)}^{2}}} \cos (ω t + Φ) + \frac{(\frac{1}{ω C} - ω L)}{\sqrt{R^{2} + {(\frac{1}{ω C} - ω L)}^{2}}} \sin (ω t + Φ)]$ $= V_{m} \cos ω t$
$\Rightarrow A \sqrt{R^{2} + {(\frac{1}{ω C} - ω L)}^{2}} [\cos α \cos (ω t + Φ) + \sin α \sin (ω t + Φ)]$ $= V_{m} \cot ω t$
where $\tan α = \frac{\frac{1}{ω C} - ω L}{R}$
$\Rightarrow A \sqrt{R^{2} + {(\frac{1}{ω C} - ω L)}^{2}} \cos (ω t + Φ - α) = V_{m} \cos ω t$
By comparing the amplitude and phase on both sides, we get,
$A = \frac{V_{m}}{\sqrt{R^{2} + {(\frac{1}{ω C} - ω L)}^{2}}}$ and $Φ = α = \tan^{-} (\frac{\frac{1}{ω C} - ω L}{R})$ .

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

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