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Chapter 31: Problem 33

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

Expert verified
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

01

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.
02

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.
03

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).
04

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).

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