An alternating voltage source V = 260 (628t) is connected across a pure inductor of 5 mH. The inductive reactance in the circuit is:

Given: V = 260 (628t), inductance L = 5 mH = 5 10^-3 H. Find: The inductive reactance XL. For a pure inductor, XL = L From the given alternating voltage expression, the angular frequency is = 628 rad/s Therefore, XL = 628 5 10^-3 XL = 3.14 Therefore, the inductive reactance is 3.14 and the correct option is A. Given: The source is of the form V = V0 ( t). Find: Identify and compute XL. Comparing V = 260 (628t) with the standard form V = V0 ( t) we get = 628 rad/s Also, L = 5 mH = 5 10^-3 H Now apply the inductive reactance relation: XL = L = 628 5 10^-3 = 3.14 Hence, the correct option is A.

An alternating voltage source V = 260 (628t) is connected across… — Solution | JEE Physics

MCQEasyJEE 2023Alternating Current Basics

An alternating voltage source $V = 260 \sin (628t)$ is connected across a pure inductor of $5 \, \text{mH}$ . The inductive reactance in the circuit is:

A
$3.14\Omega$
B
$6.28\Omega$
C
$0.5\Omega$
D
$0.318\Omega$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $V = 260 \sin(628t)$ , inductance $L = 5 \, \text{mH} = 5 \times 10^{-3} \, \text{H}$ .

Find: The inductive reactance $X_L$ .

For a pure inductor,

X_L = \omega L

From the given alternating voltage expression, the angular frequency is

\omega = 628 \, \text{rad/s}

Therefore,

X_L = 628 \times 5 \times 10^{-3}

X_L = 3.14 \, \Omega

Therefore, the inductive reactance is $3.14 \, \Omega$ and the correct option is A.

Using the AC source form

Given: The source is of the form $V = V_0 \sin(\omega t)$ .

Find: Identify $\omega$ and compute $X_L$ .

Comparing

V = 260 \sin(628t)

with the standard form

V = V_0 \sin(\omega t)

we get

\omega = 628 \, \text{rad/s}

Also,

L = 5 \, \text{mH} = 5 \times 10^{-3} \, \text{H}

Now apply the inductive reactance relation:

X_L = \omega L = 628 \times 5 \times 10^{-3} = 3.14 \, \Omega

Hence, the correct option is A.

Common mistakes

Using $X_L = 2\pi f L$ without first identifying that the given expression already provides $\omega$ directly. This is wrong because $628$ in $\sin(628t)$ is angular frequency, not frequency. Use $X_L = \omega L$ directly.
Failing to convert $5 \, \text{mH}$ into henry. This gives an answer larger by a factor of $10^3$ . Convert correctly as $5 \, \text{mH} = 5 \times 10^{-3} \, \text{H}$ before substitution.
Reading the source expression incorrectly as giving amplitude only and ignoring the coefficient of $t$ . The coefficient of $t$ is essential because it gives $\omega$ , which determines the reactance.

Practice more Alternating Current Basics questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Alternating Current Basics

Answer

Step-by-step solution

Common mistakes

Practice more Alternating Current Basics questions

Related questions