Question

Total impedance of a series LCR circuit varies with angular frequency of the AC source connected to it as shown in the graph. The quality factor Q of the series LCR circuit is

• A. 2.5
• B. 5
• C. 1
• D. 0.4

Answer 1

The Correct Option is C

Given: 
The graph of impedance (Z) vs. angular frequency (ω) for a series LCR circuit shows resonance clearly at \( \omega_0 \).
At resonance, impedance is purely resistive and minimum.

Step 1: Understanding the Quality Factor (Q)

The Quality factor \( Q \) of a series LCR circuit at resonance is defined as:

\[ Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 CR} \]

where:
\( \omega_0 \) = Resonant angular frequency,
\( L \) = Inductance,
\( C \) = Capacitance,
\( R \) = Resistance at resonance.

Step 2: Determine Quality Factor from the given graph:

Quality factor can also be graphically determined using:

\[ Q = \frac{\omega_0}{\Delta\omega} = \frac{\text{Resonant frequency}}{\text{Bandwidth (difference between frequencies at impedance = } \sqrt{2}Z_{min})} \]

From the given graph:
- Resonance occurs at \( \omega_0 = 1.0 \, \text{rad/s} \).
- At resonance, minimum impedance, \( Z_{min} = 1\,Ω \).
- Bandwidth (\( \Delta\omega \)) is the frequency range at which the impedance is \( Z = Z_{min}\sqrt{2} = 1\times\sqrt{2} \approx 1.414\,Ω \).

Observing carefully from the given graph, impedance reaches \( \sqrt{2} \) at two points:

• Lower frequency \( \omega_1 = 0.5\, \text{rad/s} \)
• Higher frequency \( \omega_2 = 1.5\, \text{rad/s} \)

Thus, bandwidth is:

\[ \Delta\omega = \omega_2 - \omega_1 = 1.5 - 0.5 = 1.0\, \text{rad/s} \]

Step 3: Calculate Quality Factor (Q):

\[ Q = \frac{\omega_0}{\Delta\omega} = \frac{1.0}{1.0} = 1 \]

Final Conclusion:
The quality factor \( Q \) of the series LCR circuit is 1.