An ac circuit contains a resistance of 1 kΩ, a capacitor of 0.1 µF and an inductor of 1 mH connected in series. The resonance frequency of the circuit is approximately: ____.

For a series R‑L‑C circuit the resonance (natural) angular frequency $\omega ₀$ is determined only by the inductance $L$ and the capacitance $C$:

$\omega ₀\; =\; \backslash frac\{1\}\{\backslash sqrt\{LC\}\}$

The corresponding resonance frequency in hertz is obtained by dividing the angular frequency by $2\pi$:

$f₀\; =\; \backslash frac\{\omega ₀\}\{2\pi \}\; =\; \backslash frac\{1\}\{2\pi \backslash sqrt\{LC\}\}$

Given values:

$L\; =\; 1 mH\; =\; 1\times 10^\{-3\}\backslash ,\{\backslash text\{H\}\}$

$C\; =\; 0.1 \mu F\; =\; 0.1\times 10^\{-6\}\backslash ,\{\backslash text\{F\}\}\; =\; 1\times 10^\{-7\}\backslash ,\{\backslash text\{F\}\}$

First compute the product $LC$:

$LC\; =\; (1\times 10^\{-3\})(1\times 10^\{-7\})\; =\; 1\times 10^\{-10\}\backslash ,\{\backslash text\{H·F\}\}$

Take the square root:

$\backslash sqrt\{LC\}\; =\; \backslash sqrt\{1\times 10^\{-10\}\}\; =\; 1\times 10^\{-5\}$

Find the angular frequency:

$\omega ₀\; =\; \backslash frac\{1\}\{\backslash sqrt\{LC\}\}\; =\; \backslash frac\{1\}\{1\times 10^\{-5\}\}\; =\; 1\times 10^\{5\}\backslash ,\{\backslash text\{rad/s\}\}$

Finally, convert to hertz:

$f₀\; =\; \backslash frac\{\omega ₀\}\{2\pi \}\; =\; \backslash frac\{1\times 10^\{5\}\}\{2\pi \}\; \approx \; \backslash frac\{1\times 10^\{5\}\}\{6.28318\}\; \approx \; 1.5915\times 10^\{4\}\backslash ,\{\backslash text\{Hz\}\}$

Thus the resonance frequency is approximately $1.59\times 10^\{4\}\backslash ,\{\backslash text\{Hz\}\}\; =\; 15.9 kHz$. The resistance of 1 kΩ does not affect the resonance frequency, only the sharpness (quality factor) of the resonance.