An A.C. circuit contains a resistor ‘R’, an inductor ‘L’ and a capacitor ‘C’ connected in series. When it is connected to an A.C. generator of fixed output voltage and variable frequency, the current in the circuit is found to be leading the applied voltage $\frac{π}{4}$ rad, when the frequency is $f_{1}$ . When the frequency of the generator increased to $' f_{2}'$ the current is found to be lagging behind the applied voltage by $\frac{π}{4}$ rad. The resonant frequency of the circuit is

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$\frac{2 f_{1} f_{2}}{f_{1} + f_{2}}$

$\frac{f_{1} f_{2}}{f_{1} + f_{2}}$

$\frac{f_{1} + f_{2}}{2}$

$\sqrt{f_{1} f_{2}}$

answer is D.

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Detailed Solution

Since the angle between the voltage and current in both the cases are 45^o.

$\begin{array}{l} 1 = \frac{2 π f_{1} L - \frac{1}{2 π f_{1} C}}{R} \\ 1 = \frac{\frac{1}{2 π f_{2} C} - 2 π f_{2} L}{R} \\ 2 π L (f_{1} + f_{2}) = \frac{1}{2 π C} (\frac{1}{f_{1}} + \frac{1}{f_{2}}) \\ 4 π^{2} LC = \frac{1}{f_{1} f_{2}} \\ Resonant frequ ency = \sqrt{f_{1} f_{2}} \end{array}$