Question

An alternating voltage source $V =260 \sin (628 t )$ is connected across a pure inductor of $5 mH$ Inductive reactance in the circuit is :

• A. $3.14 \Omega$
• B. $6.28 S$
• C. $0.318 \Omega$
• D. $0.5 \Omega$

Hint:

The inductive reactance \( X_L \) depends on the frequency of the alternating current and the inductance. It is directly proportional to both.

Answer 1

The Correct Option is A

To find the inductive reactance in the circuit, we need to use the formula for inductive reactance, which is given by:

\(X_L = \omega L\)

where:

• \(X_L\) is the inductive reactance.
• \(\omega\) is the angular frequency.
• \(L\) is the inductance.

From the problem, we have:

• Voltage source: \(V = 260 \sin(628t)\)
• \(L = 5 \text{ mH} = 5 \times 10^{-3} \text{ H}\)

The angular frequency \(\omega\) is given directly in the function as \(628 \text{ rad/s}\).

Now, substituting the values into the formula:

\(X_L = \omega L = 628 \times 5 \times 10^{-3}\)

\(X_L = 3.14 \Omega\)

Hence, the inductive reactance of the circuit is \(3.14 \Omega\).

Therefore, the correct answer is \(3.14 \Omega\), which matches the given option.