Problem 12 - Applied Physics

Question

Find the current (in amperes) in an inductive circuit where \(L=525 \mu \mathrm{H}\), \(f=25.0 \mathrm{MHz}\), and \(E=65.0 \mathrm{~V}\)

Step-by-Step Solution

Answer

The current in the circuit is approximately 0.251 A or 251 mA.

Step 1: Convert Units

Start by converting the given inductance from microhenries to henries and the frequency from megahertz to hertz. \[ L = 525 \mu \text{H} = 525 \times 10^{-6} \text{H} \]\[ f = 25.0 \text{MHz} = 25.0 \times 10^6 \text{Hz} \]

Step 2: Calculate Reactance

Use the formula for inductive reactance which is \(X_L = 2\pi f L\) to find the reactance of the inductive circuit.Substitute \(f = 25.0 \times 10^6 \text{ Hz}\) and \(L = 525 \times 10^{-6} \text{ H}\): \[ X_L = 2\pi (25.0 \times 10^6)(525 \times 10^{-6}) \]Calculate \(X_L\).

Step 3: Compute Reactance

Calculate the reactance from the previous step.\[ X_L \approx 2\pi (25.0 \times 10^6)(525 \times 10^{-6}) = 82.5 \pi \]\[ X_L \approx 259.21 \Omega \] (Using \(\pi \approx 3.14159\)).

Step 4: Find the Current

With the inductive reactance known, use Ohm's Law for AC circuits to find the current. The formula is \(I = \frac{E}{X_L}\).Substitute \(E = 65.0 \text{ V}\) and \(X_L \approx 259.21 \Omega\):\[ I = \frac{65.0}{259.21} \]Calculate the current \(I\).

Step 5: Calculate the Current

Finally, calculate \(I\) using the values we have determined:\[ I \approx \frac{65.0}{259.21} \approx 0.251 \text{ A} \] or 251 mA.

Key Concepts

Inductive Reactance

Inductive reactance plays a vital role in the analysis of AC circuits. It's essentially the opposition that an inductor offers to the flow of alternating current. Inductive reactance is denoted by \( X_L \) and is directly linked to both frequency \( f \) of the AC signal and the inductance \( L \) of the coil.
The formula to calculate inductive reactance is:

• \( X_L = 2\pi f L \)

Here, \( f \) is the frequency in hertz and \( L \) is the inductance in henries.
Understanding this concept is crucial since an increase in frequency or inductance results in higher reactance, meaning that the inductor resists the AC more strongly. This is why, in practice, inductive reactance is particularly significant in high-frequency circuits.

Ohm's Law

Ohm’s Law for AC circuits is slightly different compared to its DC counterpart. In AC circuits, resistance is not always straightforward due to the presence of inductive and capacitive elements.
For purely inductive circuits, where only an inductor and a voltage source are present, Ohm's Law is used as follows:

• \( I = \frac{E}{X_L} \)

Here, \( I \) is the current in amperes, \( E \) is the applied voltage in volts, and \( X_L \) is the inductive reactance in ohms.
This form of Ohm’s Law helps determine the current by relating it to the applied voltage and the circuit’s opposition due to inductance. It is essential for predicting how current will behave in AC circuits with inductive components.

Unit Conversion

Unit conversion is a foundational step when working with electrical components and circuits. Different components or specifications are often provided in different units, thus making conversion mandatory for consistent calculations.
For example:

• Converting microhenries to henries: \( L = 525 \mu \text{H} = 525 \times 10^{-6} \text{ H} \)
• Converting megahertz to hertz: \( f = 25.0 \text{MHz} = 25.0 \times 10^6 \text{Hz} \)

When units are standardized, calculations like inductive reactance become possible and reliable. Without proper unit conversion, results would be inaccurate, leading to potential mishaps in circuit design and analysis.

Electric Current Calculation

Calculating electric current in AC circuits involves precise formulary steps, aligning with the Ohm's Law adaptation for AC systems. In such circuits, the computation is dependent on evaluating resistances including inductive reactance.
Using the equation:

• \( I = \frac{E}{X_L} \)

We can find the current flowing through the circuit. For instance, substituting the voltage \( E = 65.0\, V \) and computed inductive reactance \( X_L \approx 259.21\, \Omega \), you find:

• \( I \approx \frac{65.0}{259.21} \approx 0.251 \text{ A} \)

Therefore, the current is approximately \( 0.251 \) A, or 251 mA. Calculating current in this structured manner ensures accurate assessments of circuit performances, assisting in effective circuit design and troubleshooting.