Question

The self-inductance of a coil with turns 50, flux 3 units and a current of 0.5A is:
(A) 75
(B) 150
(C) 300
(D) 450​

Answer 1

Answer:

The inductance is proportional to the ratio of flux to current. ... Explanation: The self inductance of a coil is given by L = Nφ/I, where N = 50, φ = 3 and I = 0.5. Thus L = 50 x 3/0.5 = 300 units.

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Answer 2

Answer:

The self-inductance of a coil is 300 i.e.option(C).

Explanation:

We will use the following relationship to solve the question,

$L=\frac{N\phi}{I}$      (1)

Where,

L=self inductance of the coil

Ф=change in flux through a coil

N=number of turns in the coil

I=current flowing through the coil

From the question we have,

The number of turns in the coil(N)=50

The flux(Ф)=3

The current flowing through the coil(I)=0.5A

By inserting the values in equation (1) we get;

$L=\frac{50\times 3}{0.5}$

$L=300$

Hence, the self-inductance of a coil is 300 i.e.option(C).

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