Practicing Success
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 |
$0.05 rad~s^{-1}$ $2 rad~s^{-1}$ $314 rad~s^{-1}$ $50 rad~s^{-1}$ |
$50 rad~s^{-1}$ |
The correct answer is Option (4) → $50 rad~s^{-1}$ Resonant angular frequency, $\omega_0=\frac{1}{\sqrt{L C}}$ $\Rightarrow \omega_0=\frac{1}{\sqrt{4 \times 100 \times 10^{-6}}}=\frac{1}{\sqrt{4 \times 10^{-4}}}=\frac{1}{2 \times 10^{-2}}$ $\Rightarrow \omega_0=50 rad~s^{-1}$ |