In R-L-C series circuit, we have same current at angular frequencies $ω_{1}$ and $ω_{2}$ The resonant frequency of circuit is

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$\frac{ω_{1}^{2}}{ω_{2}}$

$\frac{ω_{2}^{2}}{ω_{1}}$

$\sqrt{ω_{1} ω_{2}}$

$ω_{1} {+ω}_{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{V}{Z_{1}} = \frac{V}{Z_{2}} \\ z_{1} = z_{2} \\ \sqrt{R^{2} + {(ω_{1} L - \frac{1}{ω_{1} C})}^{2}} = \sqrt{R^{2} + {(ω_{2} L - \frac{1}{ω_{2} C})}^{2}} \\ {(ω_{1} L - \frac{1}{ω_{1} C})}^{2} = {(ω_{2} L - \frac{1}{ω_{2} C})}^{2} \\ (ω_{1} L - \frac{1}{ω_{1} C} + ω_{2} L - \frac{1}{ω_{2} C}) (ω_{1} L - \frac{1}{ω_{1} C} - ω_{2} L + \frac{1}{ω_{2} C}) = 0 \\ \Rightarrow ω_{1} L - \frac{1}{ω_{1} C} + ω_{2} L - \frac{1}{ω_{2} C} = 0 \\ \Rightarrow L (ω_{1} + ω_{2}) = \frac{1}{C} (\frac{1}{ω_{1}} + \frac{1}{ω_{2}}) \\ \Rightarrow ω_{1} ω_{2} = \frac{1}{L C} \\ \Rightarrow ω_{1} ω_{2} = ω_{0}^{2} \\ \Rightarrow ω_{0} = \sqrt{ω_{1} ω_{2}} \end{array}$