Question

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

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

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 \(C\): \(f = \frac{1}{2\pi (0.4\times 10^{-6})(6.63\times 10^3)}\)

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 \(59.80\ \mathrm{Hz}\).

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 \(C\): \(X_{new} = \frac{1}{2\pi (29.9)(0.4\times 10^{-6})}\)

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 \(13.26k\Omega\).