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You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.80 mA when a sinusoidal voltage with amplitude 12.0 V is applied across the inductor. What frequency is required?

Short Answer

Expert verified
The required frequency is approximately 2356.19 kHz.

Step by step solution

01

Understand the Relationship Between Voltage, Current, and Inductive Reactance

The sinusoidal voltage across an inductor is related to the current through it by the formula \( V = I \cdot X_L \), where \( V \) is the voltage amplitude, \( I \) is the current amplitude, and \( X_L \) is the inductive reactance. The inductive reactance is given by \( X_L = 2\pi f L \), where \( f \) is the frequency and \( L \) is the inductance.
02

Solve for the Inductive Reactance

Rearrange the formula \( V = I \cdot X_L \) to solve for the inductive reactance \( X_L \): \[ X_L = \frac{V}{I} \] Plug in the values: \( V = 12.0 \) V and \( I = 1.80 \) mA = 0.00180 A. \[ X_L = \frac{12.0}{0.00180} = 6666.67 \, \Omega \]
03

Solve for the Frequency

Using the formula for inductive reactance \( X_L = 2\pi f L \), solve for the frequency \( f \): \[ f = \frac{X_L}{2\pi L} \] Substituting the known values of \( X_L = 6666.67 \, \Omega \) and \( L = 0.450 \, \text{mH} = 0.000450 \, \text{H} \), \[ f = \frac{6666.67}{2\pi \cdot 0.000450} \approx 2,356,194 \, ext{Hz} \]
04

Convert the Frequency to kHz

Convert the frequency from Hertz to kilohertz by dividing by 1000: \[ f = \frac{2,356,194}{1000} \approx 2356.19 \, ext{kHz} \]

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Key Concepts

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

Inductive Reactance
When dealing with alternating current (AC) circuits containing inductors, the concept of inductive reactance is key. Inductive reactance (\(X_L\)) is the opposition that an inductor presents to a change in current. This reactance is due to the magnetic field created around the inductor. The formula to calculate inductive reactance is:
  • \(X_L = 2\pi f L\)
where:
  • \(f\) is the frequency of the sinusoidal voltage.
  • \(L\) is the inductance in henries.
Inductive reactance is measured in ohms (\(\Omega\)) and increases with either higher frequency or larger inductance. This relationship is crucial as it allows you to control how the inductor reacts to varying frequencies in a circuit.
Sinusoidal Voltage
Sinusoidal voltage is a type of alternating current (AC) that varies periodically with time, following a sine wave pattern. It is characterized by its amplitude and frequency. This makes it the primary type of AC voltage used in numerous electronic devices and electrical systems. A sinusoidal voltage can be expressed mathematically as:
  • \(V(t) = V_m \sin(2\pi ft + \phi)\)
where:
  • \(V(t)\) is the voltage at time \(t\).
  • \(V_m\) is the maximum amplitude.
  • \(f\) is the frequency.
  • \(\phi\) is the phase angle.
In this exercise, the sinusoidal voltage applied is given with an amplitude of 12.0 V, indicating the peak of the waveform. Understanding sinusoidal voltage is essential for analyzing AC circuits, as the behavior and performance of components like inductors are frequency-dependent.
Current Amplitude
Current amplitude in AC circuits refers to the maximum peak value of the current that flows through components like resistors, capacitors, or inductors. It indicates the strength of the current and is crucial for designing circuits to ensure components can handle the expected load without damage. In this exercise, the current amplitude through the inductor is specified to be 1.80 mA. This value is used with the voltage amplitude in the formula:
  • \(V = I \cdot X_L\)
where:
  • \(V\) is the voltage amplitude.
  • \(I\) is the current amplitude.
  • \(X_L\) is the inductive reactance.
By rearranging the formula, we can find the required inductive reactance and thus the frequency needed to achieve this current amplitude through the inductor, ensuring the circuit operates as intended.
Frequency Calculation
Frequency calculation in AC circuits is crucial for defining how often the voltage waveform repeats itself in a given time frame. It is measured in hertz (Hz) and plays a central role in determining how inductors and capacitors will interact within a circuit. In the context of this problem, determining the correct frequency ensures that the current amplitude of 1.80 mA is achieved under a sinusoidal voltage of 12.0 V across the inductor. The frequency can be calculated using the formula:
  • \(f = \frac{X_L}{2\pi L}\)
Here:
  • \(X_L\) is the previously calculated inductive reactance.
  • \(L\) is the inductance.
Inserting the known values allows us to find the exact frequency, which was calculated to be approximately 2,356 kHz. This value ensures the proper performance of the radio receiver circuit, showing how important accurate frequency calculations are in electronic design.

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Most popular questions from this chapter

A series circuit has an impedance of 60.0 \(\Omega\) and a power factor of 0.720 at 50.0 Hz. The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity?

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

An inductor with \(L\) = 9.50 mH is connected across an ac source that has voltage amplitude 45.0 V. (a) What is the phase angle \(\phi\) for the source voltage relative to the current? Does the source voltage lag or lead the current? (b) What value for the frequency of the source results in a current amplitude of 3.90 A?

A series ac circuit contains a 250-\(\Omega\) resistor, a 15-mH inductor, a 3.5-\(\mu\)F capacitor, and an ac power source of voltage amplitude 45 V operating at an angular frequency of 360 rad/s. (a) What is the power factor of this circuit? (b) Find the average power delivered to the entire circuit. (c) What is the average power delivered to the resistor, to the capacitor, and to the inductor?

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

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