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Chapter 30: Problem 38

A variable capacitor used in an RLC circuit produces a resonant frequency of \(5.0 \mathrm{MHz}\) when its capacitance is set to \(15 \mathrm{pF}\). What will the resonant frequency be when the capacitance is increased to \(380 \mathrm{pF} ?\)

Short Answer

Expert verified
Answer: The final resonant frequency when the capacitance is increased to 380 pF is approximately 2.04 MHz.

Step by step solution

01

Identify the formula for resonant frequency

To find the resonant frequency in an RLC circuit, we will use the formula for the resonant frequency in RLC circuits: \(f = \frac{1}{2 \pi \sqrt{LC}}\) where \(f\) is the resonant frequency, \(L\) is the inductance, and \(C\) is the capacitance.
02

Set up a proportion between initial and final conditions

Using the resonant frequency formula, we can set up a proportion between the initial and final conditions: \(\frac{f_1}{f_2} = \frac{1}{2 \pi \sqrt{L}} \cdot \frac{\sqrt{C_2}}{\sqrt{C_1}}\) where \(f_1\) is the initial resonant frequency (\(5.0 \mathrm{MHz}\)), \(f_2\) is the final resonant frequency, \(C_1\) is the initial capacitance (\(15 \mathrm{pF}\)), and \(C_2\) is the final capacitance (\(380 \mathrm{pF}\)).
03

Solve for \(f_2\)

We can now solve for \(f_2\): \(f_2 = f_1 \cdot \frac{1}{2 \pi \sqrt{L}} \cdot \frac{\sqrt{C_2}}{\sqrt{C_1}} \cdot \frac{2 \pi \sqrt{L}}{1}\) \(f_2 = f_1 \cdot \frac{\sqrt{C_2}}{\sqrt{C_1}}\) Substitute the given values: \(f_2 = 5.0 \mathrm{MHz} \cdot \frac{\sqrt{380 \mathrm{pF}}}{\sqrt{15 \mathrm{pF}}}\) \(f_2 = 5.0 \mathrm{MHz} \cdot \frac{\sqrt{380}}{\sqrt{15}}\) \(f_2 \approx 2.04 \mathrm{MHz}\)
04

State the final resonant frequency

Now we have found the final resonant frequency when the capacitance is increased to \(380 \mathrm{pF}\): \(f_2 \approx 2.04 \mathrm{MHz}\)

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Most popular questions from this chapter

A vacuum cleaner motor can be viewed as an inductor with an inductance of \(100 . \mathrm{mH} .\) For a \(60.0-\mathrm{Hz} \mathrm{AC}\) voltage, \(V_{\mathrm{rms}}=115 \mathrm{~V}\), what capacitance must be in series with the motor to maximize the power output of the vacuum cleaner?

A capacitor with capacitance \(C=5.00 \cdot 10^{-6} \mathrm{~F}\) is connected to an AC power source having a peak value of \(10.0 \mathrm{~V}\) and \(f=100 . \mathrm{Hz} .\) Find the reactance of the capacitor and the maximum current in the circuit.

A 10.0 - \(\mu\) F capacitor is fully charged by a 12.0 - \(\mathrm{V}\) battery and is then disconnected from the battery and allowed to discharge through a 0.200 - \(\mathrm{H}\) inductor. Find the first three times when the charge on the capacitor is \(80.0-\mu \mathrm{C},\) taking \(t=0\) as the instant when the capacitor is connected to the inductor.

A \(300 .-\Omega\) resistor is connected in series with a \(4.00-\mu F\) capacitor and a source of time-varying emf providing \(V_{\mathrm{rms}}=40.0 \mathrm{~V}\) a) At what frequency will the potential drop across the capacitor equal that across the resistor? b) What is the rms current through the circuit when this occurs?

A 2.00 - \(\mu\) F capacitor was fully charged by being connected to a 12.0 - \(V\) battery. The fully charged capacitor is then connected in series with a resistor and an inductor: \(R=50.0 \Omega\) and \(L=0.200 \mathrm{H}\). Calculate the damped frequency of the resulting circuit.
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