Question

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

The required frequency is approximately 2356.19 kHz.

Step-by-step solution

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.

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. \[ X_L = \frac{12.0}{0.00180} = 6666.67 \, \Omega \]

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 \( L = 0.450 \, \text{mH} = 0.000450 \, \text{H} \), \[ f = \frac{6666.67}{2\pi \cdot 0.000450} \approx 2,356,194 \, ext{Hz} \]

Convert the Frequency to kHz

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

Key concepts

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.