Question

At what frequency will a \(10.0-\mu \mathrm{F}\) capacitor have the reactance \(X_{C}=200 . \Omega ?\)

Short answer

Answer: Approximately 79.58 Hz.

Step-by-step solution

Understand the capacitive reactance formula

The formula for capacitive reactance is given by: \(X_{C} = \frac{1}{2 \pi f C}\) where \(X_{C}\) is the capacitive reactance, \(f\) is the frequency, and \(C\) is the capacitance. In this exercise, we are given \(X_{C}=200\Omega\) and \(C=10.0 \mu F\). Our task is to find the value of the frequency \(f\).

Rearrange the formula for the frequency

We need to isolate \(f\) in order to find the frequency, so we rearrange the capacitive reactance formula: \(f = \frac{1}{2 \pi X_{C}C}\)

Plug in the given values and solve for frequency

Now, we plug in the given values for \(X_{C}\) and \(C\) into the rearranged formula from Step 2: \(f = \frac{1}{2 \pi (200\Omega)(10.0\times10^{-6}F)}\) Perform the calculations: \(f = \frac{1}{2 \pi (200)(10.0\times10^{-6})}\) \(f = \frac{1}{2 \pi (2\times10^{3})(10\times10^{-6})}\) \(f \approx 79.58 Hz\)

Express the answer

At approximately 79.58 Hz, the 10.0 µF capacitor will have a reactance of 200 Ω.

Key concepts

Frequency Calculation

Understanding how to calculate the frequency at which a capacitor exhibits a specific reactance is crucial for many electronic applications. The frequency determines how a capacitive circuit responds to different signals. In our exercise, we sought the frequency at which a capacitor with a capacitance of 10 microfarads (\(10.0-\text{μF}\)) would have a reactance (\(X_{C}\)) of 200 ohms (\(200 \text{Ω}\)).

Frequency, in the context of capacitive reactance, refers to the rate at which an alternating current (AC) changes direction per second. It is measured in Hertz (Hz) and is inversely proportional to the capacitive reactance, meaning that as frequency increases, reactance decreases, and vice versa. By rearranging the reactance formula, we obtained a mathematical expression to find the frequency. This equation characterizes the inverse relationship between frequency and reactance in a capacitive circuit.

Capacitance

Capacitance is a measure of a capacitor's ability to store an electric charge per unit voltage. It is expressed in farads (F), which represents the amount of electric charge stored for a potential difference of one volt. In practice, farads are usually too large to be used for common capacitors, so microfarads (\text{μF}) and other subunits are often used.

A capacitor's capacitance depends on the physical characteristics of the device, including the area of the plates, the distance between them, and the dielectric material used. In our example, the capacitor in question has a capacitance of 10.0 microfarads (\(10.0-\text{μF}\)). Capacitance plays a direct role in determining the reactance of a capacitor at a given frequency, which brings us to the importance of the reactance formula in analyzing AC circuits.

Reactance Formula

The reactance formula for a capacitor, denoted as \(X_{C}\), is used to calculate the opposition that a capacitor presents to the flow of alternating current. It is given by the equation \(X_{C} = \frac{1}{2 \pi f C}\right), where \(X_{C}\) is the capacitive reactance in ohms, \(f\) is the frequency in Hertz (Hz), and \(C\) is the capacitance in farads (F). Note that the reactance of a capacitor decreases with increasing frequency and increases with increasing capacitance.

Equipped with the correct formula and a clear understanding of the relationship between frequency, capacitance, and reactance, solving the exercise becomes straightforward. Simply rearrange the reactance formula to solve for the unknown variable—in this case, frequency—and substitute the given values to find that at approximately 79.58 Hz, the 10.0 μF capacitor exhibits a reactance of 200 Ω. The practical implication of this calculation is fundamental for designing and troubleshooting AC circuits, particularly in tuning and filtering applications where specific reactance values are required at certain frequencies.