Question

A series LCR circuit R = 10 $k\Omega$, L = 20 mH, C = 5 $\mu$F is connected to AC source v = 10 sin$\omega$t. Find out band width of signal.

• A. 80 KHz
• B. 800 KHz
• C. 100 KHz
• D. 120 KHz

Answer 1

The correct option is A 80 KHz

Given,

$R=10k\Omega$

$L=20mH$

$C=5\mu F$

$V=10sin\omega t$

In LCR circuit,

Band width of the signal is,

Band Width, $BW=\frac{f_r}{Q}$. . . . .(1)

resonant frequency, $f_r=\frac{1}{2\pi \sqrt{LC}}$

$f_r=\frac{1}{2\times 3.14\sqrt{20\times {10}^{-3}\times 5\times {10}^{-6}}}$

$f_r=0.50KHz$

Quality factor, $Q=\frac{1}{R}\sqrt{\frac{L}{C}}$

$Q=\frac{1}{10\times {10}^3}\sqrt{\frac{20\times {10}^{-3}}{5\times {10}^{-6}}}$

$Q=0.632\times {10}^{-2}$

From equation (1)

$BW=\frac{0.50}{0.632\times {10}^{-2}}Hz$

$BW=80KHz$

The correct option is A.