Question

A series LCR circuit has R= 5$\Omega,$ L= 40mH and C=1 $\mu$F, the bandwidth of the circuit is

• A. 10Hz
• B. 20Hz
• C. 30Hz
• D. 40Hz

Answer 1

Answer: (B)

20Hz

Answer 2

: Resonant angular frequency

$\omega_{r} = \frac{1}{\sqrt{LC}}$ …(i)

Quality factor,

$Q = \frac{{Re}sonantangularfrequency}{Bandwidth}$

Bandwidth,$= \upsilon_{2} - \upsilon_{1} = \frac{\upsilon_{r}}{Q}$ …(ii)

Where, $\upsilon_{r} =$ resonant frequency$= \frac{1}{2\pi\sqrt{LC}}$

Q = quality factor.

Also, $Q = \frac{\omega_{r}L}{R}$

$\therefore\upsilon_{2} - \upsilon_{1} = \frac{\upsilon_{r}R}{2\pi\upsilon_{r}L} = \frac{R}{2\pi L}$ (Using (i) and (ii))

$\upsilon_{2} - \upsilon_{1} = \frac{R}{2\pi L} = \frac{5}{2\pi \times 40 \times 10^{- 3}} = 20Hz$