In the given AC circuit, when switch S is at position 1, the source e.m.f. leads current by $π$ /6. Now, if the switch is at position 2, then current leads source e.m.f. by $π$ /4, then :

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Frequency of a.c source is $\frac{\sqrt{2.5}}{π} kHz$

Impedence of the circuit in position 1 is 2R / 3

Impedence of the circuit in position 2 is $\sqrt{2}$ R

$R = 3 \sqrt{10} Ω$

answer is A, B, D.

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Detailed Solution

Switch at position 1
$\tan \frac{π}{6} = \frac{ωL}{R} \Rightarrow ωL = \frac{R}{\sqrt{3}}$
Switch at position 2
$\begin{array}{l} \tan \frac{π}{4} = \frac{1}{ωRC} \Rightarrow ωRC = 1 ∴ \frac{1}{ωC} = \sqrt{3} ωL \\ ω = \frac{1}{\sqrt{\sqrt{3} LC}} = \frac{1}{\sqrt{3 \times 10^{- 3} \times \frac{100}{3} \times 10^{- 6}}} \end{array} ∴ f = \frac{\sqrt{2.5}}{π} kH z$