Question

A pure inductive coil allows a current of 10A to flow from
a 230V, 50HZ supply Determine
i) Inductance of the coil​

Answer 1

The Inductance of the coil is 0.073H.

Given: A pure inductive coil having,

current = 10A,

Voltage = 230 V and

Frequency = 50 Hz.

To find: Inductance of the coil

Solution:

Inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it. L is used to represent the inductance, and Henry is the SI unit of inductance.

Inductive reactance is the opposition offered by the inductor in an AC circuit to the flow of ac current. It is represented by (XL) and measured in ohms (Ω).

Inductance is calculated as,

L = XL / 2πf

where f is the frequency.

So, we will find XL first.

XL = Ev/Iv

Here, Ev = 230 V and Iv = 10 A

XL = 230/10

XL = 23 Ω

Now,

L =  XL / 2πf

L = 23 / 2π * 50

L = 0.073 H.

Hence, Inductance of coil is L = 0.073 H.

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