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An $ac$ source of angular frequency $\omega$ is fed across a resistor r and a capacitor $C$ in series. The current registered is $I$. If now the frequency of source is changed to $\omega$ $/3$ (but maintaining the same voltage), the current in then circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency $\omega$
$\sqrt {\frac{3}{5}} $
$\sqrt {\frac{2}{5}} $
$\sqrt {\frac{1}{5}} $
$\sqrt {\frac{4}{5}} $
Solution
(a) At angular frequency $\Omega$, the current in $RC$ circuit is given by
${i_{rms}} = \frac{{{V_{rms}}}}{{\sqrt {{R^2} + {{\left( {\frac{1}{{\omega C}}} \right)}^2}} }}$ ……$(i)$
Also $\frac{{{i_{rms}}}}{2} = \frac{{{V_{rms}}}}{{\sqrt {{R^2} + {{\left( {\frac{1}{{\frac{\omega }{3}C}}} \right)}^2}} }} = \frac{{{V_{rms}}}}{{\sqrt {{R^2} + \frac{9}{{{\omega ^2}{C^2}}}} }}$ ……$(ii)$
From equation $(i)$ and $(ii)$ we get
$3{R^2} = \frac{5}{{{\omega ^2}{C^2}}} \Rightarrow \frac{{\frac{1}{{\omega C}}}}{R} = \sqrt {\frac{3}{5}} $==> $\frac{{{X_C}}}{R} = \sqrt {\frac{3}{5}} $
