See Our New JEE Book on Amazon

An RC consists of a resistor (R) and a capacitor (C) connected in series to an AC source. The behavior of the circuit depends on the values of R and C and the frequency of the applied AC voltage. At low frequencies, the capacitor behaves as an open circuit and at high frequencies, it behaves as a short circuit. The time constant $\tau=RC$ of the circuit determines the rate at which the voltage across the capacitor changes in response to changes in the AC voltage.

The capacitive impedance of the capacitor is $X_C=\frac{1}{\omega C}.$ The impedance of RC circuit is given by \begin{align} Z&=\sqrt{R^2+{X_C}^2}\\ &=\sqrt{R^2+\frac{1}{\omega^2 C^2}}. \end{align} At low frequencies, the capacitor behaves as an open circuit ($X_C\to\infty$) and most of the voltage is dropped across the resistor. At high frequencies, the capacitor behaves as a short circuit and most of the voltage is dropped across the capacitor.

The time constant of RC circuit is given by \begin{align} \tau=RC. \end{align}

In RC circuit, the current $i$ leads the voltage $V$ by a phase difference $\phi$ given by, \begin{align} \phi&=\tan^{-1}\left(\frac{X_C}{R}\right)\\ &=\tan^{-1}\left(\frac{1}{\omega RC}\right) \end{align}

**Problem (IIT JEE 2003)**
When an AC source of emf $V=V_0 \sin(100t)$ is connected across a circuit, the phase difference between the emf $V$ and the current $I$ in the circuit is observed to be $\pi/4$, as shown in the diagram. If the circuit consists possibly only RC or RL or LC in series, find the relationship between the two elements.

- $R=1\,\mathrm{k\Omega}$, $C=10\,\mathrm{\mu F}$
- $R=1\,\mathrm{k\Omega}$, $C=1\,\mathrm{\mu F}$
- $R=1\,\mathrm{k\Omega}$, $L=10\,\mathrm{H}$
- $R=1\,\mathrm{k\Omega}$, $C=1\,\mathrm{H}$

**Solution:**
From the given diagram $I$ is leading $V$, which is a characteristic of a series RC circuit. The phase difference,
\begin{align}
\tan\phi=1=\frac{X_C}{R}=\frac{1}{\omega RC},
\end{align}
gives
\begin{align}
RC=\frac{1}{\omega}=\frac{1}{100}.
\end{align}

**Problem (IIT JEE 2011)** A series RC combination is connected to an AC voltage of angular frequency $\omega=500$ rad/s. If the impedance of the $RC$ circuit is $R\sqrt{1.25}$, the time constant (in milliseconds) of the circuit is___________.

**Solution:**
The impedance of $RC$ circuit is
\begin{align}
Z&=\sqrt{R^2+\frac{1}{(\omega C)^2}}\\
&=R\sqrt{1.25}.
\end{align}
On simplification, we get time constant
\begin{align}
\tau=RC=\frac{2}{\omega}=4\times{10}^{-3}\;\mathrm{s}.
\end{align}

**Problem (IIT JEE 2010)**
An AC voltage source of variable angular frequency $\omega$ and fixed amplitude $V_0$ is connected in series with a capacitance $C$ and an electric bulb of resistance $R$ (inductance zero). When $\omega$ is increased,

- the bulb glows dimmer.
- the bulb glows brighter.
- total impedance of the circuit is unchanged.
- total impedance of the circuit increases.

**Solution:** The impedance of the RC circuit,
\begin{align}
Z=\sqrt{R^2+\frac{1}{\omega^2C^2}},
\end{align}
decreases with increase in $\omega$. This increases the current in the circuit as $I=V_0/Z$. The power of the bulb, $I^2R$, increases and hence the bulb glows brighter.

**Problem (IIT JEE 2011)** A series RC circuit is connected to AC voltage source. Consider two cases (A) when $C$ is without a dielectric medium and (B) when $C$ is filled with dielectric medium of dielectric constant $4$. The current $I_R$ through the resistor and voltage $V_C$ across the capacitor are compared in two cases. Which of the following is (are) true?

- $I_{R}^{A} > I_{R}^{B}$
- $I_{R}^{A} < I_{R}^{B}$
- $V_{C}^{A} > V_{C}^{B}$
- $V_{C}^{A} < V_{C}^{B}$

**Solution:**
Introducing the dielectric medium makes the capacitance in Case B four times of the capacitance in Case A. Thus, the circuit impedances in the two cases are given by
\begin{align}
Z^A&=\sqrt{R^2+\frac{1}{\omega^2 C^2}},
\end{align}
and
\begin{align}
Z^B&=\sqrt{R^2+\frac{1}{16\omega^2 C^2}}.
\end{align}

The currents through $R$ (and also through $C$) in the two cases are \begin{align} I_R^A &=V/Z^A \\ I_R^B &=V/Z^B \end{align} which gives $I_R^A < I_R^B$ (since $Z^A > Z^B$).

The voltages across $C$ in the two cases are \begin{align} V_C^A & =I_R^A X_C^A \\ &=\frac{V}{\sqrt{R^2+\frac{1}{\omega^2 C^2}}}\frac{1}{\omega C} \\ &=\frac{V}{\sqrt{R^2 \omega^2 C^2+1}}, \end{align} and \begin{align} V_C^B & =I_R^B X_C^B \\ &=\frac{V}{\sqrt{R^2+\frac{1}{16\omega^2 C^2}}}\frac{1}{4\omega C} \\ &=\frac{V}{\sqrt{16R^2 \omega^2 C^2+1}} \\ & < V_C^A. \end{align}