Problem

An inductor with an inductance of $100 \, \mu\mathrm{H}$ passes a current of $10 \, \mathrm{mA}$ when its terminal voltage is $6.3 \, \mathrm{V}$. Calculate the frequency of the AC supply.

Data

• Inductance: $L = 100 \, \mu\mathrm{H} = 100 \times 10^{-6} \, \mathrm{H}$
• Current: $I = 10 \, \mathrm{mA} = 10 \times 10^{-3} \, \mathrm{A}$
• Voltage: $V = 6.3 \, \mathrm{V}$

Prerequisite Concepts

1. The inductive reactance is defined as:

$$X_L = \omega L = 2 \pi f L$$

where $\omega$ is the angular frequency, and $f$ is the frequency of the AC supply. 2. Ohm’s law for inductive reactance:

$$X_L = \frac{V}{I}$$

Rearranging these equations allows us to find $f$:

$$f = \frac{X_L}{2 \pi L}$$

Solution

Step 1: Calculate the inductive reactance $X_L$

Using Ohm’s law for inductive reactance:

$$X_L = \frac{V}{I}$$

Substitute the values:

$$X_L = \frac{6.3 \, \mathrm{V}}{10 \times 10^{-3} \, \mathrm{A}}$$ $$X_L = 0.63 \times 10^3 \, \Omega = 630 \, \Omega$$

Step 2: Calculate the frequency $f$

From the formula:

$$f = \frac{X_L}{2 \pi L}$$

Substitute the values:

$$f = \frac{630 \, \Omega}{2 \pi \times 100 \times 10^{-6} \, \mathrm{H}}$$

Simplify:

$$f = \frac{630}{2 \times 3.14 \times 100 \times 10^{-6}}$$ $$f = 1.003 \times 10^6 \, \mathrm{Hz}$$

Answer

The frequency of the AC supply is:

$$f = 1.003 \, \mathrm{MHz}$$

---