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(a) What is the reactance of a \(3.00-\mathrm{H}\) inductor at a frequency of 80.0 \(\mathrm{Hz}\) ? (b) What is the inductance of an inductor whose reactance is 120\(\Omega\) at 80.0 \(\mathrm{Hz}\) ? (c) What is the reactance of a \(4.00-\mu \mathrm{F}\) capacitor at a frequency of 80.0 \(\mathrm{Hz}\) ? (d) What is the capacitance of a capacitor whose reactance is 120\(\Omega\) at 80.0 \(\mathrm{Hz} ?\)

Short Answer

Expert verified
(a) 1507.96 Ω, (b) 0.2387 H, (c) 497.42 Ω, (d) 16.57 μF.

Step by step solution

01

Formula for Inductive Reactance

The reactance of an inductor (\(X_L\)) is given by the formula \(X_L = 2\pi f L \), where \(f\) is the frequency and \(L\) is the inductance. This formula will be used for parts (a) and (b).
02

Calculate Reactance for Inductor in Part (a)

Using the formula for inductive reactance \(X_L = 2\pi (80.0 \, \text{Hz})(3.00 \, \text{H})\), compute the value: \[ X_L = 2\pi \times 80 \times 3 = 480\pi \approx 1507.96\, \Omega. \]
03

Solve for Inductance in Part (b)

Using the same formula solved for \(L\), \(L = \frac{X_L}{2\pi f}\). Substituting \(X_L = 120 \, \Omega\) and \(f = 80.0 \, \text{Hz}\): \[ L = \frac{120}{2\pi \times 80} \approx 0.2387 \, \text{H}. \]
04

Formula for Capacitive Reactance

The reactance of a capacitor (\(X_C\)) is given by the formula \(X_C = \frac{1}{2\pi f C} \), where \(f\) is the frequency and \(C\) is the capacitance. This formula will be applied for parts (c) and (d).
05

Calculate Reactance for Capacitor in Part (c)

Use \(X_C = \frac{1}{2\pi f C}\) with \(f = 80.0 \, \text{Hz}\) and \(C = 4.00 \times 10^{-6} \, \text{F}\): \[ X_C = \frac{1}{2\pi \times 80 \times 4.00 \times 10^{-6}} \approx 497.42\, \Omega. \]
06

Solve for Capacitance for Part (d)

Rearranging the formula \(X_C = \frac{1}{2\pi f C}\) to find \(C\), \(C = \frac{1}{2\pi f X_C}\), with \(X_C = 120 \, \Omega\): \[ C = \frac{1}{2\pi \times 80 \times 120} \approx 16.57 \times 10^{-6} \, \text{F} \text{ or } 16.57 \, \mu\text{F}. \]

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Key Concepts

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

Inductive Reactance
Inductive reactance is an opposition to the change of current caused by an inductor in an AC circuit. It's critical to understanding how inductors work in different electrical systems. The reactance, denoted as \(X_L\), is calculated using the formula \(X_L = 2\pi f L\). Here, \(f\) represents the frequency in hertz (Hz) and \(L\) is the inductance in henries (H).
- **Frequency Dependence**: Inductive reactance increases as the frequency increases. When frequency is higher, the inductor opposes the current more strongly. This is because a faster-changing current creates a larger opposing voltage due to the inductor's magnetic field.
- **Real-world Application**: Inductors are used in various applications such as filters and transformers. Their ability to resist changes in current flow makes them crucial for controlling alternating current characteristics.
By understanding inductive reactance, we can better design circuits that operate optimally at different frequencies.
Capacitive Reactance
Capacitive reactance is another form of reactance that occurs in AC circuits, particularly involving capacitors. This concept is essential when we want to manage how capacitors affect AC signals. The mathematical representation of capacitive reactance, \(X_C\), follows the formula \(X_C = \frac{1}{2\pi f C}\), where \(f\) is the frequency and \(C\) is the capacitance in farads (F).
- **Inversely Proportional**: Unlike inductive reactance, capacitive reactance decreases as the frequency increases. This is because capacitors reduce their opposition to current flow as the frequency of the signal rises.
- **Importance in AC Circuits**: Capacitors are often used to block direct current while allowing alternating current to pass. This quality makes capacitive reactance a vital parameter in tuning circuits like radios or in phase-shifting applications in audio equipment.
Understanding capacitive reactance lets us predict how well a capacitor will respond to different frequencies, enhancing circuit performance.
Frequency
Frequency is the rate at which an alternating current changes direction in a circuit, measured in hertz (Hz). It's a crucial factor influencing both inductive and capacitive reactance in AC circuits. Here’s why frequency is important:
- **Relation to Reactance**: Frequency directly influences reactance, determining how much opposition a component (like an inductor or capacitor) presents to the current. High frequency might significantly increase inductive reactance while decreasing capacitive reactance.
- **Circuit Design**: Engineers and circuit designers must consider frequency when selecting elements to ensure correct operation of devices. Equipment designed for low frequencies might not perform well at higher frequencies due to reactance changes.
- **Applications**: Understanding frequency allows for effective application in various areas like telecommunications, audio electronics, and power systems, where different frequency levels serve different functional roles.
By mastering the influence of frequency on circuit components, especially reactance, designers can create more efficient and effective electrical systems.

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Most popular questions from this chapter

An \(L-R-C\) series circuit draws 220 \(\mathrm{W}\) from a \(120-\mathrm{V}\) (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 capacistor 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?

A Step-Down Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply 12.0 \(\mathrm{V}(\mathrm{rms})\) to a portable electronic device. The load resistance in the secondary is 5.00\(\Omega .\) (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the \(120-\mathrm{V}\) line would draw the same power as the transformer? Show that this is equal to 5.00\(\Omega\) times the square of the ratio of primary to secondary turns.

You want to double the resonance angular frequency of an L-R-C series circuit by changing only the pertinent circuit elements all by the same factor. (a) Which ones should you change? (b) By what factor should you change them?

In an \(L-R-C\) series circuit, the source has a voltage amplitude of \(120 \mathrm{V}, R=80.0 \Omega,\) and the reactance of the capacitor is 480\(\Omega .\) The voltage amplitude across the capacitor is 360 \(\mathrm{V}\) . (a) What is the current amplitude in the circuit? (b) What is the impedance? (c) What two values can the reactance of the inductor have? (d) For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? Explain.

The power of a certain \(\mathrm{CD}\) player operating at 120 \(\mathrm{V} \mathrm{rms}\) is 20.0 \(\mathrm{W}\) . Assuming that the CD player behaves like a pure resistor, find (a) the maximum instantaneous power; (b) the rms current; \((\mathrm{c})\) the resistance of this player.

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