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Chapter 21: Problem 52

An RLC series circuit is built with a variable capacitor. How does the resonant frequency of the circuit change when the area of the capacitor is increased by a factor of \(2 ?\)

Short Answer

Expert verified
Answer: When the area of the capacitor is increased by a factor of 2, the resonant frequency of the RLC series circuit is multiplied by √(1/2), which implies that the resonant frequency decreases.

Step by step solution

01

Write down the initial capacitance and resonant frequency formulas

Initial capacitance formula: $$ C = \varepsilon_0\frac{A}{d} $$ Initial resonant frequency formula for RLC circuit: $$ f_r = \frac{1}{2 \pi \sqrt{LC}} $$
02

Express the new capacitance when the area is increased by a factor of 2

New area of the capacitor: \(A' = 2A\). Substitute this value into the capacitance formula: $$ C' = \varepsilon_0\frac{A'}{d} = \varepsilon_0\frac{2A}{d} = 2C $$ where \(C'\) denotes the new capacitance.
03

Express the new resonant frequency with the new capacitance

Now, substitute the new capacitance \(C'\) into the resonant frequency formula: $$ f_r' = \frac{1}{2 \pi \sqrt{L(2C)}} $$ where \(f_r'\) denotes the new resonant frequency.
04

Compare the initial and new resonant frequencies

We want to determine the relationship between the initial resonant frequency \(f_r\) and the new resonant frequency \(f_r'\): $$ \frac{f_r'}{f_r} = \frac{\frac{1}{2 \pi \sqrt{L(2C)}}}{\frac{1}{2 \pi \sqrt{LC}}} = \sqrt{\frac{1}{2}} $$
05

Conclude the relationship between the area and resonant frequency

The ratio of the new resonant frequency to the initial resonant frequency is \(\sqrt{\frac{1}{2}}\). Therefore, when the area of the capacitor is increased by a factor of 2, the resonant frequency of the RLC series circuit is multiplied by \(\sqrt{\frac{1}{2}}\), which implies that the resonant frequency decreases.

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