Question

Suppose an RLC circuit in resonance is used to produce a radio wave of wavelength \(150 \mathrm{~m}\). If the circuit has a 2.0 -pF capacitor, what size inductor is used?

Short answer

Answer: The inductor has a value of \(1.677 \times 10^{-6} \mathrm{H}\).

Step-by-step solution

Recall the resonance frequency formula for RLC circuits

In an RLC circuit, the resonance frequency (f) is given by the formula: \[f=\frac{1}{2 \pi \sqrt{LC}}\] Where L represents the inductance (in Henries) and C is the capacitance (in Farads).

Determine the wave frequency

Since the circuit is producing radio waves of wavelength 150 m, we need to find the frequency (f) of the wave using the speed of light (c) equation: \[c= f \times \lambda\] Where c is the speed of light (approximately \(3 \times 10^8 \mathrm{~m/s}\)), f is the frequency, and λ is the wavelength. Rearranging the equation to solve for f, we get: \[f = \frac{c}{\lambda}\]

Calculate the wave frequency

Now, plug in the values for c and λ into the formula and solve for the frequency (f): \[f = \frac{3 \times 10^8 \mathrm{~m/s}}{150 \mathrm{~m}} = 2 \times 10^6 \mathrm{Hz}\]

Calculate the inductance (L) of the inductor from the resonance frequency formula

Now, we will use the calculated frequency and given capacitor value to find the inductance (L). Rearrange the resonance frequency formula to solve for L: \[L = \frac{1}{(2 \pi f)^2 C}\]

Plug in the values and find the inductor's value

Now, substitute the calculated frequency f and given capacitance C into the formula: \[L = \frac{1}{(2 \pi \times 2 \times 10^6 \mathrm{Hz})^2 \times 2.0 \times 10^{-12} \mathrm{F}} = 1.677 \times 10^{-6} \mathrm{H}\] So, the inductor used in the RLC circuit has an inductance of \(1.677 \times 10^{-6} \mathrm{H}\).