The question asks for the resonance frequency ($f_0$) of a series AC circuit containing a resistor (R), capacitor (C), and inductor (L).
Resonance Frequency Formula
The resonance frequency in an RLC circuit is the frequency at which the inductive reactance equals the capacitive reactance. It is independent of the resistance and is calculated using the formula:
$f_0 = \frac{1}{2\pi\sqrt{LC}}$
Circuit Component Values
Given values are:
- Inductance, $L = 1 \text{ mH} = 1 \times 10^{-3} \text{ H}$
- Capacitance, $C = 0.1 \ \mu\text{F} = 0.1 \times 10^{-6} \text{ F} = 1 \times 10^{-7} \text{ F}$
- Resistance, $R = 1 \text{ k}\Omega = 1000 \ \Omega$ (Note: Resistance does not affect the resonance frequency itself).
Calculation Steps
- Substitute the values of L and C into the resonance frequency formula:
$f_0 = \frac{1}{2\pi\sqrt{(1 \times 10^{-3} \text{ H}) \times (1 \times 10^{-7} \text{ F})}}$.
- Simplify the term under the square root:
$LC = 1 \times 10^{-10} \text{ H}\cdot\text{F}$
$\sqrt{LC} = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \text{ s}$
- Calculate the resonance frequency in Hertz (Hz):
$f_0 = \frac{1}{2\pi \times (1 \times 10^{-5})} = \frac{10^5}{2\pi} \text{ Hz}$
Using $\pi \approx 3.14159$,
$f_0 \approx \frac{100000}{2 \times 3.14159} \approx \frac{100000}{6.28318} \approx 15915.5 \text{ Hz}$
- Convert the frequency to kilohertz (kHz):
$f_0 \approx 15.9155 \text{ kHz}$
Conclusion
The calculated resonance frequency is approximately $15.9 \text{ kHz}$.
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then the refractive index of the medium is ________.
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