Question

A certain series resonant circuit has a bandwidth of \[2kHz\]. If the existing coil is replaced with one having a higher value of $Q$ factor, the bandwidth will
(a) Increase
(b) Decrease
(c) Remain the same
(d) Be less selective

Answer 1

Hint: According to the relation between $Q$ factor, bandwidth is inversely proportional to the $Q$ factor. So, if the $Q$ factor increases, then bandwidth decreases. To find whether the bandwidth of the resonant circuit is increasing or decreasing, we have to find that the $Q$ factor is increasing or decreasing.

Step by step answer: In order to calculate the bandwidth, with the one having a higher value of the Q factor, we need to use the formula as mentioned below.
$Q$ factor or quality factor is related as,
\[Q=\dfrac{{{F}_{r}}}{{{B}_{w}}}\]
\[{{B}_{w}}\] \[=\]Bandwidth
\[{{F}_{r}}~=\]Resonance Frequency
So, from the above formula it can be observed as \[Q\] increases, \[{{B}_{w}}\] decreases.

Thus, the correct answer to this question is option (b).

Additional Information:The quality factor or \[Q\] factor is a dimensionless parameter that describes how underdamped an oscillator or resonator is. It is defined as the ratio of the peak energy stored in the resonator in a cycle of oscillation to the energy lost per radian of the cycle. The \[Q\] factor of the can be determined by the real part of the impedance frequency spectrum, which is defined as \[Q~=\text{ }{{f}_{r}}/\Delta fa\], where the resonance frequency \[{{f}_{r}}\] is the frequency at which the real part of the impedance reaches its maximum, \[\Delta f\] is the width of the peak at its half height, so-called \[3\text{ }dB\] bandwidth.

Note: While solving this question, we should have the basic knowledge of quality factor or \[Q\] factor. Concept of these will be used to solve this question. Some of the basic concepts are mentioned above. The formula of quality factor or \[Q\] factor used is as the \[Q\] factor of the can be determined by the real part of the impedance frequency spectrum, which is defined as \[Q~=\text{ }{{f}_{r}}/\Delta fa\].