A resonant ac circuit contains a capacitor of capacitance $10^{- 6} F$ and an inductor of $10^{- 4} H$ . The frequency of electrical oscillations will be

10^{5} H z

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10 H z

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\frac{10^{5}}{2 π} H z

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\frac{10}{2 π} H z

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Solution

The correct option is C $\frac{10^{5}}{2 π} H z$
$v = \frac{1}{2 π \sqrt{L C}} = \frac{1}{2 π \sqrt{10^{- 6} \times 10^{- 4}}} = \frac{10^{5}}{2 π} H z$ .

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