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An inductor of $0.5 mH,$ a capacitor of $200 μF$ and a resistor of $2 Ω$ are connected in series a $220 V$ AC source. If the current is in phase with the emf, the frequency of AC source will be …….. $\times 10^{2} H z .$ $(u s e, π^{2} = 10)$

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answer is 5.

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Detailed Solution

As current is in phase with emf, so the circuit is in resonance, and frequency of source=frequency of resonance

$= \frac{1}{2 π \sqrt{L C}} = \frac{1}{2 \times π \times \sqrt{0.5 \times 10^{- 3} \times 200 \times 10^{- 6}}} = 0.5 \times 10^{3} = 5 \times 10^{2} H z$

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