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31.11. Kitchen Capacitance. The wiring for a refrigerator contains a starter capacitor. A voltage of amplitude 170 \(\mathrm{V}\) and frequency 60.0 \(\mathrm{Hz}\) applied across the capacitor is to produce a current amplitude of 0.850 A through the capacitor. What capacitance \(C\) is required?

Short Answer

Expert verified
The required capacitance is approximately 13.3 μF.

Step by step solution

01

Identify Key Formula

To solve for the required capacitance, we start with the formula for capacitive reactance, which is given by \( X_C = \frac{1}{2\pi f C} \), where \( f \) is the frequency. The relationship between capacitive reactance and current is \( I = \frac{V}{X_C} \). We need to find the capacitance \( C \).
02

Calculate Capacitive Reactance

First, using the formula \( I = \frac{V}{X_C} \), we rearrange to find \( X_C = \frac{V}{I} \). Substitute the given values, \( V = 170 \, \text{V} \), and \( I = 0.850 \, \text{A} \). Thus, \( X_C = \frac{170}{0.850} \approx 200 \; \text{ohms} \).
03

Substitute into Capacitance Formula

Using the formula for capacitive reactance \( X_C = \frac{1}{2\pi f C} \), substitute the known values. We have \( X_C = 200 \), and \( f = 60.0 \, \text{Hz} \). Rearranging gives \( C = \frac{1}{2\pi \times 60 \times 200} \).
04

Calculate Capacitance

Now calculate \( C \) using the rearranged formula. Compute \( C \) as \( C = \frac{1}{2\pi \times 60 \times 200} \approx 1.33 \times 10^{-5} \, \text{F} \), or \( 13.3 \mu\text{F} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitive Reactance
Capacitive reactance is a measure of a capacitor's opposition to alternating current (AC). It is similar to resistance in a resistive circuit, but it differs in that it depends on the frequency of the applied voltage as well as on the capacitance itself. Unlike resistance, capacitive reactance decreases as the frequency increases, which is a unique property of capacitors. This property is captured in the formula:
  • \( X_C = \frac{1}{2\pi f C} \)
  • where \( X_C \) is capacitive reactance in ohms, \( f \) is the frequency in hertz, and \( C \) is the capacitance in farads.
Understanding capacitive reactance is crucial in AC circuits because it allows us to calculate how effectively a capacitor can smooth out or filter voltage fluctuations. For instance, in the exercise, we calculated the capacitive reactance to ensure the right amount of current flowed through a starter capacitor connected to a refrigerator.
AC Circuits
AC circuits are electrical circuits powered by alternating current, which reverses direction periodically. This periodic reversal is characterized by the frequency of the AC signal, commonly measured in hertz (Hz). A key component of AC circuits is the way different elements such as resistors, capacitors, and inductors react to the AC. In AC circuits, capacitors play a crucial role:
  • They store energy in the form of an electric field and release it back into the circuit, providing a stabilizing effect.
  • They also introduce a phase shift between voltage and current, which is central to their utility in tuning and timing applications.
AC circuits are foundational in the design and operation of various electrical systems, such as power grids and household appliances. In our exercise, understanding how the starter capacitor functions within the AC circuit of a refrigerator is key to determining the necessary capacitance.
Electrical Engineering Calculations
Electrical engineering calculations enable us to design and analyze electrical and electronic circuits precisely. Having a grasp on calculations involving AC circuits, capacitive reactance, and other parameters is vital. To perform these calculations effectively:
  • Identify the relationships between voltage, current, and impedance elements like resistance and reactance.
  • Apply relevant formulas, such as Ohm's Law for AC, \( V = I \times Z \), alongside specific formulas for frequency-dependent elements. In this context, \( Z \) represents the complex impedance, accounting for both resistive and reactive components.
In the given exercise, we effectively used electrical engineering calculations to determine the capacitance needed for the refrigerator's starter capacitor to function correctly. This involved using the relationship between the desired current and voltage to first find the capacitive reactance and then translating that into the required capacitance using the frequency information given.

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Most popular questions from this chapter

31.59. In an \(L . R-C\) series circuit the magnitude of the phase angle is \(54.0^{\circ}\) , with the source voltage lagging the current. The reactance of the capacitor is 350\(\Omega\) , and the resistor resistance is \(180 \Omega .\) The average power delivered by the source is 140 \(\mathrm{W}\) . Find (a) the reactance of the inductor, \((\mathrm{b})\) the rms current; (c) the rms voltage of the source.

31.62. A series circuit consists of a \(1.50-\mathrm{mH}\) inductor, a \(125-\Omega\) resistor, and a \(25.0-\mathrm{nF}\) capacitor connected across an ac source having an rms voltage of 35.0 \(\mathrm{V}\) and variable frequency. (a) At what angular frequency will the current amplitude be equal to \(\frac{1}{3}\) of its maximum possible value? (b) At the frequency in part (a) what are the current amplitude and the voltage amplitude across each of the circuit elements (including the ac source)?

31.6. A capacitance \(C\) and an inductance \(L\) are operated at the same angular frequency. (a) At what angular frequency will they have the same reactance? (b) If \(L=5.00 \mathrm{mH}\) and \(C=3.50 \mu \mathrm{F}\) , what is the numerical value of the angular frequency in part (a), and what is the reactance of each element?

31\. 72 . An \(L-R-C\) series circuit 220 \(\mathrm{W}\) from a \(120-\mathrm{V}(\mathrm{rms})\) 50.0-Hz ac line. The power factor is 0.560 , and the source voltage leads the current. (a) What is the net resistance \(R\) of the circuit? (b) Find the capacitance of the series capacitor that will result in a power factor of unity when it is added to the original circuit. (c) What power will then be drawn from the supply line?

31.28. An L-R-C series circuit is connected to a \(120-\mathrm{Hz}\) ac source that has \(V_{\mathrm{rms}}=80.0 \mathrm{V}\) . The circuit has a resistance of 75.0\(\Omega\) and an impedance at this frequency nf 105\(\Omega\) . What average power is delivered to the circuit by the source?

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