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An inductor is to be connected to the terminals of a generator (rms voltage \(=15.0 \mathrm{V})\) so that the resulting rms current will be \(0.610 \mathrm{A} .\) Determine the required inductive reactance.

Short Answer

Expert verified
The required inductive reactance is approximately 24.59 Ω.

Step by step solution

01

Understanding RMS Values

RMS (Root Mean Square) values are used to represent AC voltages and currents as if they were DC values. The given problem involves an RMS voltage of 15.0 V and an RMS current of 0.610 A. Our goal is to determine the inductive reactance that will lead to this current with the given voltage.
02

Use Ohm's Law for AC Circuits

In AC circuits with inductance or capacitance, Ohm's Law is expressed in terms of impedance (Z). For a purely inductive circuit, the impedance is the inductive reactance (X_L). Using Ohm's Law for AC circuits, we have the formula \(V_{rms} = I_{rms} \cdot X_L\), where \(V_{rms}\) is the RMS voltage and \(I_{rms}\) is the RMS current.
03

Rearrange the Formula to Isolate Inductive Reactance

To find the inductive reactance \(X_L\), rearrange the equation from Step 2: \(X_L = \frac{V_{rms}}{I_{rms}}\). This formula will allow us to calculate the value of the inductive reactance given the RMS voltage and current.
04

Substitute Known Values into the Formula

Substitute the given RMS voltage \(15.0\,V\) and RMS current \(0.610\,A\) into the formula: \(X_L = \frac{15.0\,V}{0.610\,A}\).
05

Calculate the Inductive Reactance

Perform the division to calculate \(X_L = \frac{15.0}{0.610} \approx 24.59\,\Omega\). This result gives the required inductive reactance.

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Key Concepts

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

RMS Voltage
When we talk about AC circuits, the term RMS (Root Mean Square) voltage is often used to describe the effective value of the alternating voltage. Unlike direct current (DC) where the voltage is constant, AC voltage varies with time. The RMS voltage is a way to express this varying voltage as an equivalent DC value that would deliver the same power to a load.

For example, household electricity is usually rated at an RMS voltage of about 120 V or 230 V, depending on the country. This RMS value is key because it helps in designing electrical devices to work efficiently. It's also the value you see on most meters, making it a practical measure for comparing different sources of electricity.

Calculating the RMS voltage involves taking the square root of the average of the squares of all instantaneous voltages over one cycle. While this sounds complex, it's typically simplified by formulas in practical applications.

In the problem provided, the RMS voltage is given as 15.0 V. This is the effective voltage needed for the inductor to be connected across the generator, and it helps in calculating other components like RMS current and inductive reactance.
RMS Current
RMS current, similar to RMS voltage, refers to the effective value of an alternating current (AC). It represents the equivalent direct current (DC) value that would produce the same heat in a resistor. This concept is crucial for understanding how AC circuits work, as it provides a consistent way to describe AC currents in terms that are comparable to DC.

To calculate RMS current, you may again square the instantaneous current values over a full cycle, find the average, and then take the square root of this average. However, in typical situations, especially with sinusoidal sources, it becomes straightforward with standard electrical formulas.

In this exercise, the RMS current is given as 0.610 A. This value is used along with the RMS voltage in applying Ohm's Law for AC circuits to determine other quantities, such as inductive reactance. Knowing the RMS current allows us to predict the behavior of the circuit when subjected to the RMS voltage.
Ohm's Law for AC Circuits
Ohm's Law is a fundamental principle in electronics, and it extends into AC circuits with some adjustments. In AC circuits, the standard Ohm's Law must be stated in terms of impedance due to the presence of inductors and capacitors, which introduce phase differences between voltage and current.

The law in AC form is written as: \[ V_{rms} = I_{rms} \times Z \]where \( V_{rms} \) is the RMS voltage, \( I_{rms} \) is the RMS current, and \( Z \) represents the impedance of the circuit.

For circuits with only inductors, like the exercise, the impedance is specifically the inductive reactance \( X_L \). Therefore, Ohm's Law becomes:\[ V_{rms} = I_{rms} \times X_L \]

Rearranging gives the formula used to find inductive reactance:\[ X_L = \frac{V_{rms}}{I_{rms}} \]

This relationship is fundamental in calculating the inductive reactance when both RMS voltage and current are known, as seen in this problem. By substituting the given values, we determined that the inductive reactance \( X_L \) is approximately \( 24.59 \Omega \), which helps us understand how the inductor will behave in the circuit.

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Most popular questions from this chapter

A \(30.0-\mathrm{mH}\) inductor has a reactance of \(2.10 \mathrm{k} \Omega .\) (a) What is the frequency of the ac current that passes through the inductor? (b) What is the capacitance of a capacitor that has the same reactance at this frequency? The frequency is tripled, so that the reactances of the inductor and capacitor are no longer equal. What are the new reactances of (c) the inductor and (d) the capacitor?

A capacitor is connected across the terminals of an ac generator that has a frequency of \(440 \mathrm{Hz}\) and supplies a voltage of \(24 \mathrm{V} .\) When a second capacitor is connected in parallel with the first one, the current from the generator increases by 0.18 A. Find the capacitance of the second capacitor.

A series \(\mathrm{RCL}\) circuit is composed of a resistor \((R=220 \Omega),\) two identical capacitors \((C=3.50 \mathrm{nF})\) connected in series, and two identical inductors \(\left(L=5.50 \times 10^{-5} \mathrm{H}\right)\) connected in series. You and your team need to determine: (a) the resonant frequency of this configuration. (b) What are all of the other possible resonant frequencies that can be attained by reconfiguring the capacitors and inductors (while using all of the components and keeping the proper series RCL order)? (c) If you were to design a circuit using only one of the given inductors and one adjustable capacitor, what would the range of the variable capacitor need to be in order to cover all of the resonant frequencies found in (a) and (b)?

Multiple-Concept Example 3 reviews some of the basic ideas that are pertinent to this problem. A circuit consists of a \(215-\Omega\) resistor and a \(0.200-\mathrm{H}\) inductor. These two elements are connected in series across a generator that has a frequency of \(106 \mathrm{Hz}\) and a voltage of \(234 \mathrm{V}\). (a) What is the current in the circuit? (b) Determine the phase angle between the current and the voltage of the generator.

A circuit consists of an \(85-\Omega\) resistor in series with a \(4.0-\mu \mathrm{F}\) capacitor, and the two are connected between the terminals of an ac generator. The voltage of the generator is fixed. At what frequency is the current in the circuit one-half the value that exists when the frequency is very large?

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