A 2H inductor has negligible resistance and is connected to a $\frac{50}{\pi}$ Hz A.C supply. The reactance of the inductor is?

1. Analyze the Question

We are given an inductor with an inductance ($L$) of $2\text{ H}$ connected to an A.C. supply with a frequency ($f$) of $\frac{50}{\pi}\text{ Hz}$. We need to calculate the inductive reactance ($X_L$).

2. Explain the Concept

In an A.C. circuit, an inductor opposes the flow of current. This opposition is called inductive reactance ($X_L$). The formula to calculate it is:

Where:

• $X_L$ is the inductive reactance measured in Ohms ($\Omega$).
• $\pi$ is the constant (approximately 3.142).
• $f$ is the frequency of the supply in Hertz (Hz).
• $L$ is the inductance in Henrys (H).

3. Show the Solution

Given:

• $L = 2\text{ H}$
• $f = \frac{50}{\pi}\text{ Hz}$

Substitute these values into the formula:

Now, simplify the expression:

• The $\pi$ in the numerator and the $\pi$ in the denominator cancel each other out:

4. Identify the Answer

Comparing our result of $200\Omega$ with the given options, it matches Option A.

5. Exam Tip

When dealing with A.C. circuits in JAMB Physics, always watch out for the $\pi$ in the frequency. Often, the examiners include it specifically so that it cancels out with the $\pi$ in the reactance formula ($2\pi fL$). If you see $\pi$ in the frequency, don't rush to convert it to 3.142; keep it as a symbol and look for the cancellation!