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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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Most popular questions from this chapter

An RLC series circuit is driven by a sinusoidal emf at the circuit's resonant frequency. (a) What is the phase difference between the voltages across the capacitor and inductor? [Hint: since they are in series, the same current \(i(t) \text { flows through them. }]\) (b) At resonance, the rms current in the circuit is \(120 \mathrm{mA}\). The resistance in the circuit is \(20 \Omega .\) What is the rms value of the applied emf? (c) If the frequency of the emf is changed without changing its rms value, what happens to the rms current? (W) tutorial: resonance)
An ac circuit has a single resistor, capacitor, and inductor in series. The circuit uses \(100 \mathrm{W}\) of power and draws a maximum rms current of $2.0 \mathrm{A}\( when operating at \)60 \mathrm{Hz}\( and \)120 \mathrm{V}$ rms. The capacitive reactance is 0.50 times the inductive reactance. (a) Find the phase angle. (b) Find the values of the resistor, the inductor, and the capacitor.
In an RLC circuit, these three clements are connected in series: a resistor of \(20.0 \Omega,\) a \(35.0-\mathrm{mH}\) inductor, and a 50.0 - \(\mu\) F capacitor. The ac source of the circuit has an rms voltage of \(100.0 \mathrm{V}\) and an angular frequency of $1.0 \times 10^{3} \mathrm{rad} / \mathrm{s} .$ Find (a) the reactances of the capacitor and inductor, (b) the impedance, (c) the rms current, (d) the current amplitude, (e) the phase angle, and (f) the rims voltages across each of the circuit elements. (g) Does the current lead or lag the voltage? (h) Draw a phasor diagram.
(a) What mas current is drawn by a \(4200-\mathrm{W}\) electric room heater when running on \(120 \mathrm{V}\) rms? (b) What is the power dissipation by the heater if the voltage drops to \(105 \mathrm{V}\) rms during a brownout? Assume the resistance stays the same.
A 4.00 -mH inductor is connected to an ac voltage source of \(151.0 \mathrm{V}\) rms. If the rms current in the circuit is \(0.820 \mathrm{A},\) what is the frequency of the source?
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