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A Radio Inductor. You want the current amplitude through a \(0.450-\mathrm{mH}\) inductor (part of the circuitry for a radio receiver) to be 2.60 \(\mathrm{mA}\) when a sinusoidal voltage with amplitude 12.0 \(\mathrm{V}\) is applied across the inductor. What frequency is required?

Short Answer

Expert verified
The required frequency is approximately 1633 Hz.

Step by step solution

01

Identify the given values

We are provided with the following values:- Inductance, \( L = 0.450 \) mH = \( 0.450 \times 10^{-3} \) H- Desired current amplitude, \( I = 2.60 \) mA = \( 2.60 \times 10^{-3} \) A- Voltage amplitude, \( V = 12.0 \) V.Our goal is to find the frequency \( f \) that satisfies these conditions.
02

Use Ohm's Law for Impedance

For a purely inductive circuit, the impedance \( Z \) is given by Ohm's Law as:\[Z = \frac{V}{I}\]where \( V \) is the voltage amplitude and \( I \) is the current amplitude. Substitute the given values:\[Z = \frac{12.0 \, \mathrm{V}}{2.60 \times 10^{-3} \, \mathrm{A}}\]
03

Calculate the Impedance

Calculate the impedance using the given values:\[Z = \frac{12.0}{2.60 \times 10^{-3}} = 4615.38 \, \Omega\]
04

Relate Impedance and Inductive Reactance

For an inductor, the impedance equals its inductive reactance, which is given by:\[Z = X_L = 2 \pi f L\]where \( f \) is the frequency and \( L \) is the inductance. Substitute \( Z = 4615.38 \, \Omega \) and \( L = 0.450 \times 10^{-3} \) H into this equation.
05

Solve for Frequency

Rearrange the equation for \( f \):\[f = \frac{Z}{2 \pi L}\]Substitute the known values:\[f = \frac{4615.38}{2 \pi \times 0.450 \times 10^{-3}}\]
06

Calculate the Frequency

Calculate the frequency \( f \):\[f \approx \frac{4615.38}{2 \pi \times 0.450 \times 10^{-3}} \approx 1632.99 \, \mathrm{Hz}\]
07

Final Answer

The frequency required for the current amplitude to be 2.60 mA with a 12.0 V voltage across the inductor is approximately 1633 Hz.

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle used to relate voltage, current, and resistance in electrical circuits. It states that the current through a conductor between two points is directly proportional to the voltage across the two points. In mathematical form, Ohm's Law is expressed as: \[ V = I imes R \] Here:
  • \( V \) is the voltage (in volts)
  • \( I \) is the current (in amperes)
  • \( R \) is the resistance (in ohms)
In AC circuits, specifically in the case of inductors, we use the concept of impedance \( Z \), which replaces the resistance \( R \) and accounts for the reactance due to inductors and capacitors. By extending Ohm's Law to AC circuits, it becomes: \[ Z = \frac{V}{I} \] This allows us to calculate the impedance when the voltage and current are known. It's crucial for understanding how components like inductors behave when alternating current is applied.
Frequency Calculation
The frequency of an alternating current (AC) circuit is crucial because it determines how often the current changes direction per second. It's measured in hertz (Hz). Frequency is especially important for components like inductors and capacitors, which react differently at various frequencies. To find the frequency that will produce a specific condition in a circuit, such as a desired current with a given voltage across an inductor, we can rearrange the formula for inductive reactance: \[ X_L = 2 \pi f L \] Where:
  • \( X_L \) is the inductive reactance in ohms
  • \( f \) is the frequency in hertz
  • \( L \) is the inductance in henries
By solving for \( f \), we have: \[ f = \frac{Z}{2 \pi L} \] Using this formula helps us ensure that we can control the current and other properties of an AC circuit through precise frequency adjustments.
Impedance in AC Circuits
Impedance is a crucial concept in AC circuits. It's similar to resistance in DC circuits but includes the effects of both resistance and reactance (from inductors and capacitors). In mathematical terms, impedance is represented as \( Z \) and is measured in ohms (\( \Omega \)). Impedance can be thought of as the total opposition to the current flow in an AC circuit. It's made up of two components:
  • Resistive component \( (R) \)
  • Reactive component which includes inductive \( (X_L) \) and capacitive reactance \( (X_C) \)
For purely inductive circuits, impedance \( Z \) is equal to inductive reactance \( X_L \). This is given by: \[ Z = X_L \] In these circuits, impedance increases with frequency. That's why high-frequency signals face more opposition in inductors compared to low-frequency signals. Understanding impedance is essential for designing and troubleshooting circuits in various applications, ensuring that they perform efficiently in their intended frequency range.

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Most popular questions from this chapter

A \(100-\Omega\) resistor, a \(0.100-\mu \mathrm{F}\) capacitor, and a \(0.300-\mathrm{H}\) inductor are connected in parallel to a voltage source with amplitude 240 \(\mathrm{V}\) (a) What is the resonance angular frequency? (b) What is the maximum current through the source at the resonance frequency? (c) Find the maximum current in the resistor at resonance. (d) What is the maximum current in the inductor at resonance? (e) What is the maximum current in the branch containing the capacitor at resonance? (f) Find the maximum energy stored in the inductor and in the capacitor at resonance.

A Step-Up Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply \(13,000 \mathrm{V}(\mathrm{rms})\) for a neon sign. To reduce shock hazard, a fuse is to be inserted in the primary circuit; the fuse is to blow when the rms current in the secondary circuit; exceeds 8.50 \(\mathrm{mA}\) . (a) What is the ratio of secondary to primary turns of the transformer? (b) What power must be supplied to the transformer when the rms secondary current is 8.50 \(\mathrm{mA}\) ? (c) What current rating should the fuse in the primary circuit have?

In an \(L-R-C\) series circuit, the components have the following values: \(L=20.0 \mathrm{mH}, C=140 \mathrm{nF},\) and \(R=350 \Omega .\) The generator has an rms voltage of 120 \(\mathrm{V}\) and a frequency of 1.25 \(\mathrm{kHz}\) . Determine (a) the power supplied by the generator and (b) the power dissipated in the resistor.

(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 \(\mathrm{Hz} ?\)

An \(L-R-C\) series circuit with \(L=0.120 \mathrm{H}, R=240 \Omega\) and \(C=7.30 \mu\) F carries an rms current of 0.450 \(\mathrm{A}\) with a frequency of 400 \(\mathrm{Hz}\) (a) What are the phase angle and power factor for this circuit? (b) What is the impedance of the circuit? (c) What is the rms voltage of the source? (d) What average power is delivered by the source? (e) What is the average rate at which electrical energy is converted to thermal energy in the resistor? (f) What is the average rate at which electrical energy is dissipated (converted to other forms) in the capacitor? (g) In the inductor?

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