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Question

An inductance coil has a reactance of 100Ω. When an AC signal of frequency1000Hz is applied to the coil, the applied voltage leads the current by 450. The self-inductance of the coil is:


A

6.7×10-7H

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B

5.5×10-5H

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C

1.1×10-1H

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D

1.1×10-2H

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Solution

The correct option is D

1.1×10-2H


Step 1: Given data and assumption.

Impudence of the coil, Z=100Ω

Frequency, f=1000Hz

Current leads the voltage by an angle, ϕ=450

Step 2: Finding the self-inductance L of the coil.

Since, It is clear from the given data that the inductance is not pure inductance. The given coil behaves as a series combination of L and R.

Therefore,

The applied voltage leads the current by 450

tanϕ=XLR

Where, XLis inductive reactance.

tan450=XLR

1=XLR

XL=R ……i

Now,

We know that

Impedance of the coil,

Z=R2+XL2

Z=XL2+XL2 Fromequationi

100=2XL2

XL=1002 …..ii

As we know that,

XL=Lω

where ω is angular frequency

Lω=1002

L=1002×π×f×2 ω=2πf

L=1002×3.14×1000×2

L=1.1×10-2H

Hence, option D is correct.


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