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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Alternating Current
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An LCR circuit has $L= 3.0 H, C= 27 \;\mu F$ and $ R= 7.4 \;\Omega$ with a certain a factor. To improve the sharpness of resonance by reducing its full width at half maximum by a factor of 2, with out changing $w_0$, we can

$(a)\;\text{ Double the value of R} \\(b)\; \text{Double the value of C} \\ (c)\; \text{reduce the value of R to R/2} \\ (d)\;\text{Charge the value of C to C/2} $

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$W_0=\large\frac{1}{\sqrt {LC}}=\frac{1}{\sqrt {3 \times 27 \times 10^{-6}}}$
$\qquad= \large\frac{1}{9 \times 10^{-3}}\;$$rad/s$
$Q= \large\frac{w_0 L}{R}$
To double Q factor with out changing $W_0$ we can reduce $ R$ to $R/2$
Hence c is the correct answer.
answered Mar 25, 2014 by meena.p

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