Question

Suppose the reading of voltmeter $V$ as shown in the diagram at resonance is $200V$, then what will be the quality factor of the circuit?

$\begin{align}
  & A.2 \\
 & B.4 \\
 & C.1 \\
 & D.3 \\
\end{align}$

Answer 1

Hint: The voltage across the capacitor in the resonant condition will provide a relation between voltage across the capacitor and source voltage which is found by taking the product of the quality factor and the voltage of the source. Substitute the values in it and find out the answer. This will help you to solve this question.

Complete step by step answer:

The voltage across the capacitor in the resonant condition will provide a relation between voltage across the capacitor and source voltage which is found by taking the product of the quality factor and the voltage of the source. This can be written as,
${{V}_{C}}=Q\times {{V}_{S}}$
Where $Q$ be the quality factor of the circuit and ${{V}_{S}}$ be the voltage of the source.
It has been given in the question that the voltage of the source is obtained from the circuit as,
${{V}_{S}}=50V$
The voltage of the capacitor can be shown as,
${{V}_{c}}=200V$
Rearranging the equation in terms of the quality factor is given as,
$Q=\dfrac{{{V}_{C}}}{{{V}_{S}}}$
Substituting the values in it will give,
$Q=\dfrac{200}{50}=4$
The quality factor has been obtained as option B.

Note:
Quality factor is a dimensionless quantity. This has been used to explain how an underdamped an oscillator or resonator will be. It has been explained as the peak energy stored in the resonator in a cycle of oscillation which is divided by the energy lost per radian of the cycle. They are found to be inversely proportional to the quality of the coil and are calculated from the ratio of the reactance to the ohmic resistance.