In this article, we will discuss an AC circuit that contains a resistor and capacitor in series, so let’s get started…

## AC circuit containing $R$ and $C$ in series

In the below figure, let’s consider a resistor $R$ and an inductor $L$ connected in series to a source of alternating EMF $\mathcal{E}$ that is given by $$\mathcal{E}=\mathcal{E_0}\sin\omega t$$

Let $I$ be the current through the series circuit at any instant, then

1). Voltage $\overrightarrow{V_R}=\overrightarrow{I}R$ across the resistance $R$ will be in phase with the current $\overrightarrow{I}$. So phasors $\overrightarrow{V_R}$ and $\overrightarrow{I}$ are in the same direction, as shown in the below figure. The amplitude of $\overrightarrow{V_R}$ is $${V_0}^R=I_0 R$$

2). Voltage $\vec{V}_C=X_C \vec{I}$ across the capacitance $C$ lags behind the current $\vec{I}$ in phase by $\pi / 2$ rad. So phasor $\vec{V}_{\mathrm{C}}$ lies $\pi / 2 \mathrm{rad}$ clockwise w.r.t. the phasor $\vec{I}$, It amplitude is

$$V_0{ }^C=I_0 X_C $$

where $X_C$ is the capacitive reactance.

By the **parallelogram law of vector addition**,

$$\vec{V}_R+\vec{V}_C=\vec{\mathcal{E}}$$**Using the Pythagoras theorem**, we get

\begin{aligned} \mathcal{E_0}^2 &=\left(V_0^R\right)^2+\left(V_0^C\right)^2=\left(I_0 R\right)^2+\left(I_0 X_C\right)^2 \\ &=I_0^2\left(R^2+X_C^2\right) \\ I_0 &=\frac{\mathcal{E_0}}{\sqrt{R^2+X_C^2}} \end{aligned} |

**Read Also**

- AC circuit containing an inductor only
- AC circuit containing resistor only, class 12
- Representation of AC current and voltage by phasor diagram
- AC circuit containing resistor and inductor in series

## Impedance of $R-C$ circuit

Clearly, $\sqrt{R^2+X_C^2}$ is the effective resistance of the series RC-circuit which opposes or impedes the flow of a.c. through it. It is called its impedance and is denoted by $Z$. Thus

$$Z=\sqrt{R^2+X_C^2}=\sqrt{R^2+\frac{1}{\omega^2 C^2}} \quad\left[\because X_C=\frac{1}{\omega C}\right]$$ |

The phase angle $\phi$ between the resultant voltage and current is given by

\begin{aligned}\tan \phi=\frac{V_0^{C}}{V_0^R}&=\frac{I_0 X_C}{I_0 R}=\frac{X_C}{R}=\frac{1/\omega C}{R}\\ \text{Or}\quad \cos\phi =\frac{R}{Z}\end{aligned} |

It is obvious from the phasor diagram that the current is ahead of the emf by phase angle $\phi$ so the instantaneous value of current is given by

$$\left.I=I_0 \sin (\omega t+\phi\right)$$

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