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Chapter 21: Problem 15

A \(0.400-\mu \mathrm{F}\) capacitor is connected across the terminals of a variable frequency oscillator. (a) What is the frequency when the reactance is \(6.63 \mathrm{k} \Omega ?\) (b) Find the reactance for half of that same frequency.

Short Answer

Expert verified
Based on the given capacitance value and the reactance, we calculated the frequency to be approximately 59.80 Hz. When we found the reactance for half of this frequency (29.9 Hz), it came out to be approximately 13.26 kΩ.

Step by step solution

01

Determine the frequency for the given reactance

First, we have to determine the frequency when the reactance is \(6.63k\Omega\). The given reactance is \(X = 6.63\times 10^3\ \Omega\), and the capacitance value is \(C = 0.4\times 10^{-6}\ \mathrm{F}\). Recall the formula for reactance: \(X = 1 / (2 * \pi * f * C)\) We can rearrange the equation to solve for frequency \(f\): \(f = 1 / (2*\pi * C * X)\) Now, substitute the values of \(X\) and \(C\): \(f = \frac{1}{2\pi (0.4\times 10^{-6})(6.63\times 10^3)}\)
02

Calculate the frequency

Now we can calculate the frequency: \(f = \frac{1}{2\pi (0.4\times 10^{-6})(6.63\times 10^3)} \approx 59.80\ \mathrm{Hz}\) Thus, the frequency when the reactance is \(6.63k\Omega\) is approximately \(59.80\ \mathrm{Hz}\).
03

Find the reactance for half of the obtained frequency

Now we need to find the reactance for half of the obtained frequency. Let's use the reactance formula again: \(X_{new} = 1/(2 * \pi * f_{new} * C)\) But this time, the new frequency value \(f_{new}\) will be half of the obtained frequency in step 2: \(f_{new} = 59.80\ \mathrm{Hz} / 2 = 29.9\ \mathrm{Hz}\) Now, substitute the values of \(f_{new}\) and \(C\): \(X_{new} = \frac{1}{2\pi (29.9)(0.4\times 10^{-6})}\)
04

Calculate the new reactance

Now we can calculate the new reactance: \(X_{new} = \frac{1}{2\pi (29.9)(0.4\times 10^{-6})} \approx 13.26\ k\Omega\) Thus the reactance of the capacitor for half of the obtained frequency (29.9 Hz) is approximately \(13.26k\Omega\).

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Most popular questions from this chapter

A \(300.0-\Omega\) resistor and a 2.5 - \(\mu\) F capacitor are connected in series across the terminals of a sinusoidal emf with a frequency of $159 \mathrm{Hz}$. The inductance of the circuit is negligible. What is the impedance of the circuit?
A 4.0 -k \(W\) heater is designed to be connected to a \(120-V\) ms source. What is the power dissipated by the heater if it is instead connected to a \(120-\mathrm{V}\) de source?
A \(0.250-\mu \mathrm{F}\) capacitor is connected to a \(220-\mathrm{V}\) rms ac source at \(50.0 \mathrm{Hz}\) (a) Find the reactance of the capacitor. (b) What is the rms current through the capacitor?
A solenoid with a radius of \(8.0 \times 10^{-3} \mathrm{m}\) and 200 turns/cm is used as an inductor in a circuit. When the solenoid is connected to a source of \(15 \mathrm{V}\) ms at \(22 \mathrm{kHz}\), an \(\mathrm{rms}\) current of \(3.5 \times 10^{-2} \mathrm{A}\) is measured. Assume the resistance of the solenoid is negligible. (a) What is the inductive reactance? (b) What is the length of the solenoid?
A certain circuit has a \(25-\Omega\) resistor and one other component in series with a \(12-\mathrm{V}\) (rms) sinusoidal ac source. The rms current in the circuit is 0.317 A when the frequency is \(150 \mathrm{Hz}\) and increases by \(25.0 \%\) when the frequency increases to \(250 \mathrm{Hz}\). (a) What is the second component in the circuit? (b) What is the current at $250 \mathrm{Hz} ?$ (c) What is the numerical value of the second component?
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