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.

Answer 2

The quality factor \(Q\) of a series LCR circuit is given by the formula: \[ Q = \dfrac{\omega_0}{\Delta \omega} \] where \(\omega_0\) is the angular frequency at resonance and \(\Delta \omega\) is the bandwidth of the resonance. From the graph, we can observe that:
\(\omega_0\) corresponds to the angular frequency at the resonance point, which is where the impedance is minimum.
\(\Delta \omega\) is the difference between the angular frequencies at which the impedance is \(\sqrt{2} Z_{\text{min}}\), which corresponds to the points where the impedance starts increasing significantly on either side of resonance. From the graph, we can deduce that: \[ Q = \dfrac{\omega_0}{\Delta \omega} = 1 \]

Thus, the correct answer is (C) 1.