If the resistance in a parallel resonant circuit is reduced then what will happen to its bandwidth?
A) Increases
B) Decreases
C) Disappears
D) Becomes sharper

Hint: The bandwidth of an L-C-R circuit is defined as the range of frequencies for which the current or voltage of the circuit is 70.7 percent as that at the resonant frequency. Bandwidth depends on the quality factor and resonant frequency of the circuit.

Formula used:
The bandwidth of a parallel resonant circuit is given by, $Bandwidth = \dfrac{{{\omega _o}}}{Q}$, where ${\omega _o}$= Resonant frequency of the circuit, and Q =quality factor of the circuit
Quality factor is given by, $Q = \dfrac{1}{R}\sqrt {\dfrac{L}{C}} $ where, L = Inductance of the circuit, C = Capacitance of the circuit and R = Resistance of the circuit

Complete step by step solution:
First, we will obtain the formula of bandwidth by substituting the values of resonant frequency (${\omega _o}$) and quality factor (Q) in terms of inductance (L), capacitance (C), and resistance (R).
As we know that
Resonant frequency $\left( {{\omega _o}} \right) = \dfrac{1}{{\sqrt {LC} }}$
So now by substituting these values in the formula of the bandwidth, We get
$Bandwidth = \dfrac{{\dfrac{1}{{\sqrt {LC} }}}}{{\dfrac{1}{R}\sqrt {\dfrac{L}{C}} }}$
$ \Rightarrow Bandwidth = \dfrac{R}{L}$
Hence bandwidth is directly proportional to resistance(R) so when R decreases bandwidth also decreases.

Additional information:
Bandwidth of the circuit is of highly used in the working of radios and filters. Essentially when we change a radio channel we are usually changing the bandwidth of the consisting circuit which then allows a certain range of frequencies and rejects others and hence we switch to another radio channel.

Hence, The correct option is (B).

Note:
1. Bandwidth of the parallel resonant circuit is independent of the capacitance of the circuit.
2. By changing resistance( R), bandwidth changes but resonant frequency remains the same.