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.
A series LCR circuit has resonant frequency of $\omega_0$. If inductance in the circuit is replaced by another one having the same value of inductance but higher value of internal resistance.
Select the statement that explains the change in functioning of the circuit correctly.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Quality factor of the circuit will decrease
When inductance in the circuit is replaced by another one having the same value of inductance but higher of internal resistance then,
$\omega_0=\frac{1}{\sqrt{L C}}$, no effect on resonant frequency
$i_{\text {rms }}=\frac{i_0}{\sqrt{2}} \Rightarrow i_0=\sqrt{2} \times \frac{V_{rms}}{R}$ or resistance of the circuit increases hence amplitude of current at resonant frequency decreases.
Quality factor $=\frac{1}{R} \sqrt{\frac{L}{C}}$ or $R \uparrow$, quality factor $\downarrow$
Band width, $\Delta \omega=\frac{R}{2 L}$, or $R \uparrow, \Delta \omega \uparrow$