Question

An inductor of inductance 1.0 H in an A.C. circuit has a reactance of 500 ohm. The frequency of A.C. will be:​

Answer 1

Given

An inductor of inductance 1.0 H in an A.C. circuit has a reactance of 500 Ohm

To Find

The frequency of A.C.

Solution

We know that

$\sf \longrightarrow f=\dfrac{X_{L}}{2 \pi L}$

Here

• f = Frequency
• $\sf X_{L}$ = Inductive Reactance
• L = Inductance

Units

• f = Hertz (Hz)
• $\sf X_{L}$ = Ohm (Ω)
• L = Henry (H)

According to the question

We are asked to find frequency of A.C.

Therefore

We must find "f"

Given that

An inductor of inductance 1.0 H in an A.C. circuit has a reactance of 500 Ohm

Hence

• $\sf X_{L}$ = 500 Ω
• L = 1.0 H

We know that

• $\sf \pi$ = 3.14

Substituting the values

We get

$\sf \longrightarrow f=\dfrac{500}{2 \times 3.14 \times 1} \ Hz$

$\sf \longrightarrow f=\dfrac{500}{6.28} \ Hz$

$\sf \longrightarrow f=79.6178 \ Hz$

Therefore

$\sf \longrightarrow f=79.62 \ Hz$

Hence

The frequency of A.C. = 79.62 Hz

Answer 2

Hello !

Given:-

• Inductance of indicator = 1 H
• inductive reactance of conductor = 5000 ohm.

To find:-

• frequency of alternating current

Knowledge required:-

$\huge \fbox \blue{XL= ωL = 2πfL}$

Solution:-

$\implies$ $X_{L}$ = 2πfL

$\implies$ 500 = 2(3.14)f(1)

$\implies$ 500 = 6.28f

$\implies$ f = $\dfrac{500}{6.28}$

$\implies$ f = 79.62 Hz.