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Question

Physics Question on AC Voltage

The resonance frequency of the tank circuit of an oscillator when L=10π2mHL=\frac{10}{{{\pi }^{2}}}mH and C=0.04 μFC=0.04\text{ }\mu F are connected in parallel is

A

250 kHz

B

25 kHz

C

2.5 kHz

D

25 MHz

Answer

25 kHz

Explanation

Solution

In parallel resonant circuit resonance frequency fo=12πLC=12π10×103π2×0.04×106{{f}_{o}}=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{\frac{10\times {{10}^{-3}}}{{{\pi }^{2}}}\times 0.04\times {{10}^{-6}}}} =1042×0.2=25kHz=\frac{{{10}^{4}}}{2\times 0.2}=25\,kHz