
Define the quality factor of resonance in the series LCR circuit. What is its SI unit?
Answer
537.9k+ views
Hint: First understand the LCR circuit. Then get an idea about the quality factor. Also, define the quality factor using the mathematical equation. Now, using the mathematical expression find the SI unit of the same.
Complete step by step answer:
LCR circuit is a series or a parallel connection, consisting of inductor, capacitor, and resistance. If we are talking about resonance conditions, then it means that the magnitude of the inductor and capacitor is equal in the given circuit. If the magnitude of the inductor and capacitor is equal then they cancel each other’s effect and we are only left with resistance. Given below is the figure of an LCR circuit.
The quality factor is the ratio of maximum energy stored in a capacitor or inductor to the energy dissipated by a resistor or we can say energy dissipated during one cycle of oscillation. We can mathematically express it as:
$Q=\dfrac{V_{L}}{V_{R}}=\dfrac{I \times X_{L}}{I \times R}=\dfrac{\omega \times L}{R}$
where,
Q denotes quality factor,
R is the resistance in a circuit,
$V_L$ is potential across the inductor,
$V_R$ is the potential across resistor
$X_L$ is inductor impedance
$\omega$ is angular frequency
L is the inductance.
From the above expression, we can analyze that the Quality factor is unit-less quantity. Thus, it has no units.
Note: Do not forget about two important conditions given in the question. First is that the circuit is in resonance and second is that the inductor, capacitor and resistor are connected in series. Also, note that in the solution we are using potential across the inductor, but you can also use potential across the capacitor.
Complete step by step answer:
LCR circuit is a series or a parallel connection, consisting of inductor, capacitor, and resistance. If we are talking about resonance conditions, then it means that the magnitude of the inductor and capacitor is equal in the given circuit. If the magnitude of the inductor and capacitor is equal then they cancel each other’s effect and we are only left with resistance. Given below is the figure of an LCR circuit.

The quality factor is the ratio of maximum energy stored in a capacitor or inductor to the energy dissipated by a resistor or we can say energy dissipated during one cycle of oscillation. We can mathematically express it as:
$Q=\dfrac{V_{L}}{V_{R}}=\dfrac{I \times X_{L}}{I \times R}=\dfrac{\omega \times L}{R}$
where,
Q denotes quality factor,
R is the resistance in a circuit,
$V_L$ is potential across the inductor,
$V_R$ is the potential across resistor
$X_L$ is inductor impedance
$\omega$ is angular frequency
L is the inductance.
From the above expression, we can analyze that the Quality factor is unit-less quantity. Thus, it has no units.
Note: Do not forget about two important conditions given in the question. First is that the circuit is in resonance and second is that the inductor, capacitor and resistor are connected in series. Also, note that in the solution we are using potential across the inductor, but you can also use potential across the capacitor.
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