Question

Frequency of L-C circuit is ${f_1}$. If a resistance $R$ is also added to it, the frequency becomes ${f_2}$. The ratio of $\dfrac{{{f_2}}}{{{f_1}}}$ will be
(A) $\sqrt {1 + \dfrac{{{R^2}C}}{{4L}}} $
(B) $\sqrt {1 - \dfrac{{{R^2}C}}{{4L}}} $
(C) $\sqrt {1 + \dfrac{{{R^2}C}}{L}} $
(D) $\sqrt {1 - \dfrac{{{R^2}C}}{L}} $

Answer 1

Hint
This problem is related to the circuits; we must know the frequency formula ${f_1}$ for L-C circuit and also know the frequency formula ${f_2}$ when the resistance $R$ is added. This type of electric circuit comes under the LCR circuit and the LCR circuit comes under alternating current topics.

Complete step by step solution
The frequency of L-C circuit is,
${f_1} = \dfrac{1}{{\sqrt {LC} }}\,......................\left( 1 \right)$
Where,$L$ is the inductor, $C$ is the capacitor.
Now, if a resistance $R$ is added in the L-C circuit, then the frequency ${f_2}$ of the L-C circuit is written as,
${f_2} = \sqrt {{{\left( {\dfrac{1}{{\sqrt {LC} }}} \right)}^2} - {{\left( {\dfrac{R}{{2L}}} \right)}^2}} \,...............\left( 2 \right)$
Where, $R$ is the resistance added to the L-C circuit
Now, taking the ratio of $\dfrac{{{f_2}}}{{{f_1}}}$, then,
$\dfrac{{{f_2}}}{{{f_1}}} = \dfrac{{\sqrt {{{\left( {\dfrac{1}{{\sqrt {LC} }}} \right)}^2} - {{\left( {\dfrac{R}{{2L}}} \right)}^2}} }}{{\left( {\dfrac{1}{{\sqrt {LC} }}} \right)}}$
By squaring the first term inside the root in numerator, then the above equation is written as,
$\dfrac{{{f_2}}}{{{f_1}}} = \dfrac{{\sqrt {\left( {\dfrac{1}{{LC}}} \right) - {{\left( {\dfrac{R}{{2L}}} \right)}^2}} }}{{\left( {\dfrac{1}{{\sqrt {LC} }}} \right)}}$
By taking square root in denominator, then the above equation is written as,
$\dfrac{{{f_2}}}{{{f_1}}} = \sqrt {\dfrac{{\left( {\dfrac{1}{{LC}}} \right) - {{\left( {\dfrac{R}{{2L}}} \right)}^2}}}{{\left( {\dfrac{1}{{LC}}} \right)}}} $
By squaring the term $\left( {\dfrac{R}{{2L}}} \right)$ in the numerator, then the above equation is written as,
 $\dfrac{{{f_2}}}{{{f_1}}} = \sqrt {\dfrac{{\left( {\dfrac{1}{{LC}}} \right) - \left( {\dfrac{{{R^2}}}{{4{L^2}}}} \right)}}{{\left( {\dfrac{1}{{LC}}} \right)}}} $
By separating the numerator with denominator for further simplification, then the above equation is written as,
$\dfrac{{{f_2}}}{{{f_1}}} = \sqrt {\dfrac{{\left( {\dfrac{1}{{LC}}} \right)}}{{\left( {\dfrac{1}{{LC}}} \right)}} - \dfrac{{\left( {\dfrac{{{R^2}}}{{4{L^2}}}} \right)}}{{\left( {\dfrac{1}{{LC}}} \right)}}} $
By cancelling the same terms in the above equation, then,
$\dfrac{{{f_2}}}{{{f_1}}} = \sqrt {1 - \dfrac{{\left( {\dfrac{{{R^2}}}{{4{L^2}}}} \right)}}{{\left( {\dfrac{1}{{LC}}} \right)}}} $
On further simplifications, then the above equation is written as,
$\dfrac{{{f_2}}}{{{f_1}}} = \sqrt {1 - \left( {\dfrac{{{R^2}LC}}{{4{L^2}}}} \right)} $
And by cancelling the same terms, then the above equation is written as,
$\dfrac{{{f_2}}}{{{f_1}}} = \sqrt {1 - \left( {\dfrac{{{R^2}C}}{{4L}}} \right)} $
Thus, the ratio of their frequencies is, $\sqrt {1 - \left( {\dfrac{{{R^2}C}}{{4L}}} \right)} $
Hence, the option (B) is the correct answer.

Note
While solving this problem, we have to give more concentration in cancelling the same terms in numerator and denominator. Before the final step, we have to give more concentration while taking the denominator value to the numerator.