Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Current Electricity

Question:

A series LCR circuit connected to an AC source with voltage of the source $v=v_m \sin \omega t$.

If 'q' is the charge on the capacitor and 'i' is the current, from Kirchoff's loop rule:

$L \frac{d i}{d t}+i R+\frac{q}{c}=V$

The current in the circuit is given by $i=I_{m} \sin (\omega t+ \phi)$ where $\phi$ is the phase difference between the voltage across the source and current in the circuit.

We know $V_{R m}=L_m R ; V_{L m}=L_m X_L ; V_{C m}=L_m X_C$; and $X_L=\omega L ; X_C=\frac{1}{\omega C}$

Total impedance in the circuit regulates current. At resonance frequency of the LCR circuit current in the circuit is maximum.

In a series LCR circuit $L=4 H, C=100 \mu F$ and $R=25 \Omega$. The circuit is connected to a variable frequency 220 V source.

The source angular frequency $(\omega)$ which drives the circuit in resonance is

Options:

$0.05 rad~s^{-1}$

$2 rad~s^{-1}$

$314 rad~s^{-1}$

$50 rad~s^{-1}$

Correct Answer:

$50 rad~s^{-1}$

Explanation:

The correct answer is Option (4) → $50 rad~s^{-1}$

Resonant angular frequency, $\omega_0=\frac{1}{\sqrt{L C}}$

$\Rightarrow \omega_0=\frac{1}{\sqrt{4 \times 100 \times 10^{-6}}}=\frac{1}{\sqrt{4 \times 10^{-4}}}=\frac{1}{2 \times 10^{-2}}$

$\Rightarrow \omega_0=50 rad~s^{-1}$