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

The size of the inductor used in the RLC circuit in resonance that produces a radio wave with a wavelength of 150 meters and with a capacitor value of 2 pF is approximately 15.9 nH (nanohenries).

Step-by-step solution

1. Find the resonant frequency of the RLC circuit

The wavelength (λ) of the radio wave is given as 150 m. The speed of a radio wave is also the speed of light (c) in a vacuum, which is approximately \(3 \times 10^8\) m/s. We can find the resonant frequency (f) using the formula: \[f = \frac{c}{λ}\] Where, f = resonant frequency c = speed of light λ = wavelength.

2. Calculate the resonant frequency

Plug in the given values: \[f=\frac{3\times10^8\,\text{m/s}}{150\,\text{m}}\] Calculate the value: \[f=2\times10^6\,\text{Hz}\] So, the resonant frequency of the RLC circuit is 2 MHz.

3. Apply the resonance condition for the RLC circuit

In a resonant RLC circuit, the inductive reactance (X_L) equals the capacitive reactance (X_C). We can write this as: \[X_L = X_C\] Where, X_L = inductive reactance X_C = capacitive reactance The reactance formulas are given as: \[X_L = 2\pi fL\] \[X_C = \frac{1}{2\pi fC}\] Here, f = resonant frequency L = inductor size C = capacitor value (in Farads).

4. Solve for the inductor size (L)

Combine the reactance formulas and solve for L: \[2\pi fL= \frac{1}{2\pi fC}\] Rearrange the equation: \[L = \frac{1}{(2\pi f)^2 C}\] Plug in the values for f and C (note: 2 pF = 2 × 10^(-12) F): \[L= \frac{1}{(2\pi(2\times10^6)^2(2\times10^{-12}))}\] Calculate the value of L: \[L\approx 1.59\times10^{-8}\, \text{H}\] So, the inductor used in the RLC circuit is approximately \(1.59\times10^{-8}\) H (15.9 nH).