Question
Calculate the resonant frequency of a circuit of negligible resistance containing an inductance of $40.0 \mathrm{mH}$ and a capacitance of $600 \mathrm{pF}$$$f_{0}=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{\left(40.0 \times 10^{-3} \mathrm{H}\right)\left(600 \times 10^{-12} \mathrm{~F}\right)}}=32.5 \mathrm{kHz}$$
Step 1
The inductance L is given as 40.0 mH, which is equal to $40.0 \times 10^{-3}$ H. The capacitance C is given as 600 pF, which is equal to $600 \times 10^{-12}$ F. Show more…
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