Talha's Physics Academy
Flow of AC Through RC and RL Series Circuits
Video Lecture
Watch the complete video lecture below to understand alternating current flowing through RC and RL series circuits, phase relationships, impedance, and phasor diagrams.
Describe the Flow of AC Through RC Series Circuits
Consider a circuit containing a resistor ($R$) and a capacitor ($C$) connected in series with an alternating voltage source as shown in the figure. Due to the series connection, the same current will flow through each circuit element. The voltage across the resistor will be $V_R$, and the voltage across the capacitor will be $V_C$.
The phasor diagram is drawn taking current as the reference direction. As we know that the voltage lags the current by $\frac{\pi}{2}$ radians in the case of a capacitor, it is drawn perpendicular to the current phasor in the figure. In an RC series circuit, the voltage and current are in phase when considering the resistive component of the circuit ($V_R$).
From the phasor diagram, using Pythagoras' theorem:
$$V^2 = V_R^2 + V_C^2$$
$$V = \sqrt{V_R^2 + V_C^2}$$
As we know that $V_R = IR$ and $V_C = IX_C$, substituting these values gives:
$$V = \sqrt{(IR)^2 + (IX_C)^2}$$
$$V = I \sqrt{R^2 + X_C^2}$$
Rearranging for impedance ($Z = \frac{V}{I}$):
$$\frac{V}{I} = \sqrt{R^2 + X_C^2}$$
$$Z = \sqrt{R^2 + X_C^2}$$
Where $X_C = \frac{1}{2\pi f C}$ represents the capacitive reactance.
In this case, the phase angle ($\theta$) between current and voltage is given by:
$$\tan\theta = \frac{V_C}{V_R} \quad \text{or} \quad \tan\theta = \frac{X_C}{R}$$
$$\theta = \tan^{-1}\left(\frac{X_C}{R}\right)$$
Describe the Flow of AC Through RL Series Circuits
Let us consider a circuit containing an inductor ($L$) and a resistor ($R$) connected in series with an alternating voltage source as shown in the figure. The voltages across the resistor and inductor are $V_R$ and $V_L$, respectively.
When an inductor is connected to an alternating voltage, due to the generation of back EMF, the current lags the voltage by $90^\circ$ (or voltage leads current by $90^\circ$). This is represented by a line perpendicular to the reference phasor (current $I$) as shown in the figure. In the resistor, the current and voltage are in phase ($V_R$).
To calculate the impedance of the RL circuit, apply Pythagoras' theorem to the phasor diagram:
$$V^2 = V_R^2 + V_L^2$$
$$V = \sqrt{V_R^2 + V_L^2}$$
As we know that $V_R = IR$ and $V_L = IX_L$, substituting these values gives:
$$V = \sqrt{(IR)^2 + (IX_L)^2}$$
$$V = I \sqrt{R^2 + X_L^2}$$
Rearranging for impedance ($Z = \frac{V}{I}$):
$$\frac{V}{I} = \sqrt{R^2 + X_L^2}$$
$$Z = \sqrt{R^2 + X_L^2}$$
Where $X_L = 2\pi f L$ represents the inductive reactance.
In this case, the phase angle ($\theta$) between current and voltage is given by:
$$\tan\theta = \frac{V_L}{V_R} \quad \text{or} \quad \tan\theta = \frac{X_L}{R}$$
$$\theta = \tan^{-1}\left(\frac{X_L}{R}\right)$$
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