Question

An RLC circuit is driven by a sinusoidal emf with angular frequency $\omega$. If the angular frequency is increased without changing the amplitude of the emf, the current amplitude increases, it means?

• A. $\omega L>\frac{1}{\omega C}$
• B. $\omega L<\frac{1}{\omega C}$
• C. $\omega L=\frac{1}{\omega C}$
• D. None of the options

Answer 1

The correct option is C

$\omega L=\frac{1}{\omega C}$

Step 1: Given data

It is given that the angular frequency is increased without changing the amplitude of the emf and the current amplitude increases.

Step 2: Formula used

$E=IZ$

Where $E$ is emf, $I$ is current and $Z$ is the impedance of the circuit.

$$\begin{gathered} Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2} \\ Z=\sqrt{R^2+{\left(L\omega -\frac{1}{C\omega }\right)}^2} \end{gathered}$$

Where $R$ is resistance, $X_L$ is the inductive reactance, $X_C$ is the capacitive reactance, $L$ is the inductance, $C$ is the capacitance and $\omega$ is the angular frequency.

Step 3: Calculation

We know in an RLC circuit, $E=IZ$ where $E$ is emf, $I$ is current and $Z$ is the impedance of the circuit.

Here, as it is given that the amplitude of emf is constant, the current is inversely proportional to the impedance of the circuit.

$I\propto \frac{1}{Z}$

It means as current increases $Z$ must decrease. Since impedance,

$$\begin{gathered} Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2} \\ Z=\sqrt{R^2+{\left(L\omega -\frac{1}{C\omega }\right)}^2} \end{gathered}$$

$Z$ will decrease if $R=0$ or $L\omega -\frac{1}{C\omega }=0$.

As R cannot be zero,

$$\begin{gathered} \omega L-\frac{1}{\omega C}=0 \\ \Rightarrow \omega L=\frac{1}{\omega C} \end{gathered}$$

The solution is $\omega L=\frac{1}{\omega C}$

hence, option C is correct.